public class Solution
    {
        public int[] SortArrayByParityII(int[] A)
        {
            var len = A.Length;
            int[] ODD = new int[len / 2];//奇数1,3,5,7,9
            int[] EVEN = new int[len / 2];//偶数0,2,4,6,8
            int x = 0;
            int y = 0;
            for (int i = 0; i < len; i++)
            {
                if (A[i] % 2 == 0)
                {
                    EVEN[x] = A[i];
                    x++;
                }
                else
                {
                    ODD[y] = A[i];
                    y++;
                }
            }
            x = 0;
            y = 0;
            for (int i = 0; i < len; i++)
            {
                if (i % 2 == 0)
                {
                    A[i] = EVEN[x];
                    x++;
                }
                else
                {
                    A[i] = ODD[y];
                    y++;
                }
            }
            return A;
        }
    }

 

posted on 2018-10-14 19:43  Sempron2800+  阅读(114)  评论(0编辑  收藏  举报