922_Sort_Array_By_Parity_II

qiyuangong · qiyuangong · commit cd1cb103d347 · 2018-11-30T22:45:16.000+08:00
diff --git a/README.md b/README.md
@@ -160,6 +160,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 819 | [Most Common Word](https://leetcode.com/problems/most-common-word/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/819_Most_Common_Word.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/819_Most_Common_Word.java) | String processing, be careful about 'b,b,b'. regex is recommended. |
 | 844 | [Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/844_Backspace_String_Compare.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/844_Backspace_String_Compare.java) | 1. Stack pop when encounters #, O(n) and O(n)<br>2. Compare string from end to start, O(n) and O(1) |
 | 904 | [Fruit Into Baskets](https://leetcode.com/problems/fruit-into-baskets/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/904_Fruit_Into_Baskets.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/904_Fruit_Into_Baskets.java) | 1. Scan through blocks of tree, O(n) and O(n)<br>2. Mainten a sliding window with start and curr point, O(n) and O(n).  |
+| 922 | [Sort Array By Parity II](https://leetcode.com/problems/sort-array-by-parity-ii/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/922_Sort_Array_By_Parity_II.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/922_Sort_Array_By_Parity_II.java) | 1. Place odd and even number in odd and even place, not sort is needed. O(n) and O(1)<br>2. Two points with quick sort swap idea, O(n) and O(1). |
 | 929 | [Unique Email Addresses](https://leetcode.com/problems/unique-email-addresses/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/929_Unique_Email_Addresses.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/929_Unique_Email_Addresses.java) | String handle and hash (or set) |
 | 945 | [Minimum Increment to Make Array Unique](https://leetcode.com/problems/minimum-increment-to-make-array-unique/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/945_Minimum_Increment_to_Make_Array_Unique.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/945_Minimum_Increment_to_Make_Array_Unique.java) | Sort, then list duplicate and missing value in sorted list. O(nlgn) and O(n) |
 | 946 | [Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/946_Validate_Stack_Sequences.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/946_Validate_Stack_Sequences.java) | Add a stack named inStack to help going through pushed and popped. O(n) and O(n) |
diff --git a/java/922_Sort_Array_By_Parity_II.java b/java/922_Sort_Array_By_Parity_II.java
@@ -0,0 +1,32 @@
+class Solution {
+    /*public int[] sortArrayByParityII(int[] A) {
+        int N = A.length;
+        int[] ans = new int[N];
+        int t = 0;
+        for (int x: A) if (x % 2 == 0) {
+            ans[t] = x;
+            t += 2;
+        }
+        t = 1;
+        for (int x: A) if (x % 2 == 1) {
+            ans[t] = x;
+            t += 2;
+        }
+        return ans;
+    }*/
+    public int[] sortArrayByParityII(int[] A) {
+        int j = 1;
+        for (int i = 0; i < A.length; i += 2)
+            if (A[i] % 2 == 1) {
+                while (A[j] % 2 == 1)
+                    j += 2;
+
+                // Swap A[i] and A[j]
+                int tmp = A[i];
+                A[i] = A[j];
+                A[j] = tmp;
+            }
+
+        return A;
+    }
+}
diff --git a/python/922_Sort_Array_By_Parity_II.py b/python/922_Sort_Array_By_Parity_II.py
@@ -0,0 +1,27 @@
+class Solution(object):
+    # def sortArrayByParityII(self, A):
+    #     N = len(A)
+    #     ans = [None] * N
+    #     t = 0
+    #     for i, x in enumerate(A):
+    #         if x % 2 == 0:
+    #             ans[t] = x
+    #             t += 2
+    #     t = 1
+    #     for i, x in enumerate(A):
+    #         if x % 2 == 1:
+    #             ans[t] = x
+    #             t += 2
+    #     # We could have also used slice assignment:
+    #     # ans[::2] = (x for x in A if x % 2 == 0)
+    #     # ans[1::2] = (x for x in A if x % 2 == 1)
+    #     return ans
+
+    def sortArrayByParityII(self, A):
+        odd = 1
+        for i in xrange(0, len(A), 2):
+            if A[i] % 2:
+                while A[odd] % 2:
+                    odd += 2
+                A[i], A[odd] = A[odd], A[i]
+        return A
