905. Sort Array By Parity

QUESTION:

Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Note:

  1. 1 <= A.length <= 5000
  2. 0 <= A[i] <= 5000

EXPLANAITION:

其实这道题就是双指针问题。

用一个指针来表示偶数，一个指针来表示奇数。

一个从头开始往后寻找，一个从后往前寻找。

找到对应的之后，就将两者进行对调。

这样就结束了循环。

SOLUTION:

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int evenNumber = 0;
        int oddNumber = A.length-1;
        while (evenNumber<oddNumber){
            while ((A[evenNumber]&1)!=1 && evenNumber<A.length-1) evenNumber++;
            while ((A[oddNumber]&1)==1 && oddNumber>0) oddNumber--;
            if(evenNumber<oddNumber){
                swap(A,evenNumber,oddNumber);
                evenNumber++;oddNumber--;
            }
        }
        return A;
    }
    public static void swap(int[] A,int a,int b){
        int tmp = A[a];
        A[a] = A[b];
        A[b] = tmp;
    }
}
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