Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.
Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.
You may return any answer array that satisfies this condition.
Example 1:
Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
Note:
2 <= A.length <= 20000A.length % 2 == 00 <= A[i] <= 1000
class Solution { public int[] sortArrayByParityII(int[] A) { if(A==null||A.length ==0){ return null; } int[] B = new int[A.length]; int k = 0; int j =1; for(int i = 0; i<A.length; i++ ){ if(A[i]%2 ==0){ B[k] = A[i]; k = k+2; } else{ B[j] = A[i]; j = j+2; } } return B; } }
(Easy) Sort Array By Parity II - LeetCode
原文：https://www.cnblogs.com/codingyangmao/p/11459258.html