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lucifer
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feat: cpp code
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‎problems/19.removeNthNodeFromEndofList.md‎

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## 题目地址(19. 删除链表的倒数第N个节点)
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## 题目地址(19. 删除链表的倒数第 N 个节点)
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https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/
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@@ -64,7 +64,7 @@ https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/
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## 代码
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代码支持: JS, Java
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代码支持: JS, Java，CPP
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Javascript Code:
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}
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```
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CPP Code:
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```cpp
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class Solution {
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public:
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ListNode* removeNthFromEnd(ListNode* head, int n) {
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ListNode *p = head, *q = head;
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while (n--) q = q->next;
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if (!q) {
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head = head->next;
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delete p;
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return head;
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}
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while (q->next) p = p->next, q = q->next;
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q = p->next;
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p->next = q->next;
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delete q;
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return head;
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}
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};
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```
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**复杂度分析**
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- 时间复杂度：$$O(N)$$
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- 空间复杂度：$$O(1)$$
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‎problems/21.merge-two-sorted-lists.md‎

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@@ -44,7 +44,32 @@ https://leetcode-cn.com/problems/merge-two-sorted-lists
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## 代码
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- 语言支持：JS
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- 语言支持：CPP，JS
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CPP Code:
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```cpp
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class Solution {
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public:
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ListNode* mergeTwoLists(ListNode* a, ListNode* b) {
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ListNode head, *tail = &head;
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while (a && b) {
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if (a->val <= b->val) {
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tail->next = a;
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a = a->next;
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} else {
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tail->next = b;
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b = b->next;
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}
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tail = tail->next;
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}
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tail->next = a ? a : b;
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return head.next;
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}
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};
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```
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JS Code:
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```js
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/**

‎problems/22.generate-parentheses.md‎

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@@ -41,7 +41,7 @@ https://leetcode-cn.com/problems/generate-parentheses
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不难想到， 如果左括号的数目小于右括号，我们可以提前退出，这就是这道题的剪枝。 比如 `())....`，后面就不用看了，直接退出即可。回溯的退出条件也不难想到，那就是：
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- 左括号数目等于右括号数目
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- 左括号数目 + 右括号数目 = 2 * n
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- 左括号数目 + 右括号数目 = 2 \* n
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由于我们需要剪枝， 因此必须从左开始遍历。（WHY？）
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## 代码
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- 语言支持：JS，Python3
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- 语言支持：JS，Python3，CPP
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JS Code:
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@@ -134,12 +134,42 @@ class Solution:
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return res
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```
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CPP Code:
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```cpp
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class Solution {
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private:
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vector<string> ans;
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void generate(int leftCnt, int rightCnt, string &s) {
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if (!leftCnt && !rightCnt) {
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ans.push_back(s);
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return;
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}
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if (leftCnt) {
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s.push_back('(');
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generate(leftCnt - 1, rightCnt, s);
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s.pop_back();
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}
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if (rightCnt > leftCnt) {
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s.push_back(')');
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generate(leftCnt, rightCnt - 1, s);
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s.pop_back();
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}
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}
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public:
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vector<string> generateParenthesis(int n) {
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string s;
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generate(n, n, s);
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return ans;
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}
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};
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```
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**复杂度分析**
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- 时间复杂度：O(2^N)
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- 空间复杂度：O(2^N)
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大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：https://github.com/azl397985856/leetcode 。 目前已经 37K star 啦。
144174
大家也可以关注我的公众号《力扣加加》带你啃下算法这块硬骨头。
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![](https://tva1.sinaimg.cn/large/007S8ZIlly1gfcuzagjalj30p00dwabs.jpg)
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![](https://tva1.sinaimg.cn/large/007S8ZIlly1gfcuzagjalj30p00dwabs.jpg)

‎problems/23.merge-k-sorted-lists.md‎

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@@ -57,7 +57,7 @@ https://leetcode-cn.com/problems/merge-k-sorted-lists/
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## 代码
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代码支持 JavaScript， Python3
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代码支持 JavaScript， Python3, CPP
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JavaScript Code：
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return res.next
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```
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CPP Code:
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```cpp
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class Solution {
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private:
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ListNode* mergeTwoLists(ListNode* a, ListNode* b) {
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ListNode head(0), *tail = &head;
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while (a && b) {
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if (a->val < b->val) { tail->next = a; a = a->next; }
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else { tail->next = b; b = b->next; }
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tail = tail->next;
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}
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tail->next = a ? a : b;
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return head.next;
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}
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public:
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ListNode* mergeKLists(vector<ListNode*>& lists) {
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if (lists.empty()) return NULL;
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for (int N = lists.size(); N > 1; N = (N + 1) / 2) {
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for (int i = 0; i < N / 2; ++i) {
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lists[i] = mergeTwoLists(lists[i], lists[N - 1 - i]);
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}
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}
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return lists[0];
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}
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};
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```
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**复杂度分析**
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- 时间复杂度：$$O(kn*logk)$$
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- [88.merge-sorted-array](./88.merge-sorted-array.md)
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## 扩展
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这道题其实可以用堆来做，感兴趣的同学尝试一下吧。
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大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：https://github.com/azl397985856/leetcode 。 目前已经 37K star 啦。
186218
大家也可以关注我的公众号《力扣加加》带你啃下算法这块硬骨头。
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![](https://tva1.sinaimg.cn/large/007S8ZIlly1gfcuzagjalj30p00dwabs.jpg)

‎problems/24.swapNodesInPairs.md‎

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@@ -41,7 +41,7 @@ https://leetcode-cn.com/problems/swap-nodes-in-pairs/
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## 思路
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设置一个 dummy 节点简化操作，dummy next 指向 head。
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设置一个 dummy 节点简化操作，dummy next 指向 head。
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1. 初始化 first 为第一个节点
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2. 初始化 second 为第二个节点
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## 代码
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- 语言支持：JS，Python3, Go, PHP
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- 语言支持：JS，Python3, Go, PHP, CPP
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JS Code:
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* @param {ListNode} head
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* @return {ListNode}
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*/
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var swapPairs = function(head) {
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var swapPairs = function (head) {
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const dummy = new ListNode(0);
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dummy.next = head;
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let current = dummy;
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while (current.next != null && current.next.next != null) {
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// 初始化双指针
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const first = current.next;
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const second = current.next.next;
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// 更新双指针和 current 指针
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first.next = second.next;
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second.next = first;
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}
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```
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CPP Code:
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```cpp
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class Solution {
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public:
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ListNode* swapPairs(ListNode* head) {
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ListNode h, *tail = &h;
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while (head && head->next) {
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auto p = head, q = head->next;
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head = q->next;
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q->next = p;
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tail->next = q;
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tail = p;
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}
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tail->next = head;
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return h.next;
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}
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};
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```
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**复杂度分析**
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- 时间复杂度：$$O(N)$$
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- 空间复杂度：$$O(1)$$
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