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‎README.md‎

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@@ -2035,4 +2035,242 @@ public class Solution {
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//return Math.pow(x, n);
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}
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}
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```
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## 51. N-Queens
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题目：经典的N皇后问题，给出一个NxN的棋盘,求所有皇后排列的组合。
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!(8-queens)[https://leetcode.com/static/images/problemset/8-queens.png]
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要保证皇后不能互相攻击，就必须要求任意两个皇后不在同一行，同一列和同一对角线上。
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```
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For example,
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There exist two distinct solutions to the 4-queens puzzle:
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[
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[".Q..", // Solution 1
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"...Q",
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"Q...",
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"..Q."],
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["..Q.", // Solution 2
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"Q...",
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"...Q",
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".Q.."]
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]
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```
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一个NP问题，开一个n大的数组，下标表示行值表示列。每加入一个值时先与之前放入数组的值进行判断是否符合规则，符合规则继续下一个。
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```java
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public class Solution {
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private List<List<String>> resultList = new ArrayList<>();
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private int[] positions;
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private char[] tags;
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public List<List<String>> solveNQueens(int n) {
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positions = new int[n];
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tags = new char[n];
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Arrays.fill(positions, -1);
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Arrays.fill(tags, '.');
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dp(0, n);
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return resultList;
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}
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public void dp(int n, int max) {
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if (n == max) {
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List<String> qLine = new ArrayList<>();
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for (int i = 0; i < positions.length; i++) {
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tags[positions[i]] = 'Q';
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String q = new String(tags);
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qLine.add(q);
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tags[positions[i]] = '.';
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}
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resultList.add(qLine);
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return;
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}
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int valueX = n;
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int valueY = 0;
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while (valueY < max) {
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if (hasPlaced(n, valueX, valueY)) {
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positions[n] = valueY;
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dp(n + 1, max);
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}
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valueY++;
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}
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}
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public boolean hasPlaced(int n, int valueX, int valueY) {
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for (int j = 0; j < n; j++) {
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int x = j;
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int y = positions[j];
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if (positions[j] == valueY || Math.abs(valueX - x) == Math.abs(valueY - y)) {
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return false;
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}
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}
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return true;
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}
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}
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```
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## 52. N-Queens II
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与上一题一样的题目，不过是求组合的个数而不需要将组合罗列出来
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```java
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public class Solution {
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private int count = 0;
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private int[] positions;
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public int totalNQueens(int n) {
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positions = new int[n];
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Arrays.fill(positions, -1);
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dp(0, n);
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return count;
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}
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public void dp(int n, int max) {
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if (n == max) {
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count++;
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return;
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}
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int valueX = n;
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int valueY = 0;
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while (valueY < max) {
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if (hasPlaced(n, valueX, valueY)) {
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positions[n] = valueY;
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dp(n + 1, max);
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}
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valueY++;
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}
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}
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public boolean hasPlaced(int n, int valueX, int valueY) {
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for (int j = 0; j < n; j++) {
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int x = j;
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int y = positions[j];
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if (positions[j] == valueY || Math.abs(valueX - x) == Math.abs(valueY - y)) {
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return false;
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}
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}
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return true;
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}
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}
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```
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## 53. Maximum Subarray
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题目：给出一个数组，求一个和最大的子数组。
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```
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For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
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the contiguous subarray [4,-1,2,1] has the largest sum = 6.
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```
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需要一个额外的数组来维护每个位置的最大值，以便进行下一个位置最大值的计算。
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```java
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public class Solution {
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public int maxSubArray(int[] nums) {
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int[] total = new int[nums.length];
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total[0] = nums[0];
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int totalMax = total[0];
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for (int i = 1; i < nums.length; i++) {
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total[i] = Math.max(nums[i], nums[i] + total[i - 1]);
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totalMax = Math.max(totalMax, total[i]);
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}
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return totalMax;
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}
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}
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```
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## 54. Spiral Matrix
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题目：给出一组m x n 的二维数字数组，返回螺旋顺序的数字列表。
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```
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Given the following matrix:
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[
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[ 1, 2, 3 ],
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[ 4, 5, 6 ],
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[ 7, 8, 9 ]
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]
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You should return [1,2,3,6,9,8,7,4,5].
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```
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将螺旋的顺序分解开来，可以分成4种顺序，依次为从左向右，从上向下，从左向右，从下向上。定义4个变量分别为`m-剩下的行数`,`n-剩下的列数`,`startM-行开始的坐标`,`startN-列开始的地方`。
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接下来就是按顺序求列表了。
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```java
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public class Solution {
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public List<Integer> spiralOrder(int[][] matrix) {
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List<Integer> s = new ArrayList<>();
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int m = matrix.length;
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if(m == 0) return s;
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int n = matrix[0].length;
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int startM = 0, startN = 0;
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while (m > 0 && n > 0) {
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spiral1(startM, startN, m, n, matrix, s);
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startM++;
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m--;
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if (m <= 0 || n <= 0) return s;
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spiral2(startM, startN, m, n, matrix, s);
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n--;
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if (m <= 0 || n <= 0) return s;
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spiral3(startM, startN, m, n, matrix, s);
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m--;
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if (m <= 0 || n <= 0) return s;
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spiral4(startM, startN, m, n, matrix, s);
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startN++;
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n--;
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}
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return s;
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}
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public void spiral1(int startM, int startN, int m, int n, int[][] matrix, List<Integer> s) {
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for (int k = startN; k < n + startN; k++) {
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s.add(matrix[startM][k]);
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}
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}
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public void spiral2(int startM, int startN, int m, int n, int[][] matrix, List<Integer> s) {
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for (int k = startM; k < m + startM; k++) {
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s.add(matrix[k][n + startN - 1]);
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}
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}
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public void spiral3(int startM, int startN, int m, int n, int[][] matrix, List<Integer> s) {
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for (int k = startN + n - 1; k >= startN; k--) {
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s.add(matrix[startM + m - 1][k]);
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}
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}
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public void spiral4(int startM, int startN, int m, int n, int[][] matrix, List<Integer> s) {
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for (int k = m + startM - 1; k >= startM; k--) {
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s.add(matrix[k][startN]);
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}
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}
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}
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```
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## 55. Jump Game
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题目：跳棋，给出一组数字，每个数字代表当前可以跳过的最大距离，求是否能够从第一个位置达到最后一个位置。
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```
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For example:
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A = [2,3,1,1,4], return true.
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A = [3,2,1,0,4], return false.
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```
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要先记录每个位置能够跳到的最远距离，特殊处理当数值为0时能够跳过去。
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```java
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public class Solution {
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public boolean canJump(int[] nums) {
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if (nums.length <= 1) return true;
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if (nums[0] == 0) return false;
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int[] maxs = new int[nums.length];
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maxs[0] = nums[0];
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for (int i = 1; i < nums.length - 1; i++) {
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if (nums[i] == 0) {
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if (maxs[i - 1] <= 1) {
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return false;
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}
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maxs[i] = maxs[i - 1] - 1;
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}
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maxs[i] = Math.max(maxs[i - 1] - 1, nums[i]);
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}
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return true;
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}
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}
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```

‎src/JumpGame.java‎

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/**
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* Created by cpacm on 2017/6/26.
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*/
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public class JumpGame {
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public static void main(String[] args) {
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System.out.println(canJump(new int[]{0, 1}));
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}
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public static boolean canJump(int[] nums) {
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if (nums.length <= 1) return true;
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if (nums[0] == 0) return false;
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int[] maxs = new int[nums.length];
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maxs[0] = nums[0];
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for (int i = 1; i < nums.length; i++) {
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if (nums[i] == 0) {
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if (maxs[i - 1] <= 1) {
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return i >= nums.length - 1;
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}
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maxs[i] = maxs[i - 1] - 1;
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}
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maxs[i] = Math.max(maxs[i - 1] - 1, nums[i]);
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}
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return true;
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}
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}

‎src/MaxSubArray.java‎

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import java.util.Arrays;
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/**
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* Created by cpacm on 2017/6/26.
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*/
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public class MaxSubArray {
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public static void main(String[] args) {
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System.out.println(maxSubArray(new int[]{-2, 1, -3, 4, -1, 2, 1, -5, 4}));
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}
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public static int maxSubArray(int[] nums) {
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int[] total = new int[nums.length];
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total[0] = nums[0];
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int totalMax = total[0];
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for (int i = 1; i < nums.length; i++) {
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total[i] = Math.max(nums[i], nums[i] + total[i - 1]);
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totalMax = Math.max(totalMax, total[i]);
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}
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return totalMax;
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}
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}

‎src/NQueens.java‎

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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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/**
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* Created by cpacm on 2017/6/24.
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*/
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public class NQueens {
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private static List<List<String>> resultList = new ArrayList<>();
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private static int[] positions;
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private static char[] tags;
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public static void main(String[] args) {
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System.out.println(solveNQueens(8));
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}
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public static List<List<String>> solveNQueens(int n) {
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positions = new int[n];
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tags = new char[n];
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Arrays.fill(positions, -1);
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Arrays.fill(tags, '.');
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dp(0, n);
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return resultList;
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}
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public static void dp(int n, int max) {
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if (n == max) {
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List<String> qLine = new ArrayList<>();
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for (int i = 0; i < positions.length; i++) {
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tags[positions[i]] = 'Q';
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String q = new String(tags);
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qLine.add(q);
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tags[positions[i]] = '.';
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}
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resultList.add(qLine);
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return;
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}
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int valueX = n;
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int valueY = 0;
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while (valueY < max) {
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if (hasPlaced(n, valueX, valueY)) {
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positions[n] = valueY;
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dp(n + 1, max);
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}
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valueY++;
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}
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}
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public static boolean hasPlaced(int n, int valueX, int valueY) {
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for (int j = 0; j < n; j++) {
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int x = j;
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int y = positions[j];
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if (positions[j] == valueY || Math.abs(valueX - x) == Math.abs(valueY - y)) {
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return false;
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}
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}
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return true;
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}
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}

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