LEETCODE41905. Sort Array By Parity
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package y2019.Algorithm.array; /** * @ProjectName: cutter-point * @Package: y2019.Algorithm.array * @ClassName: SortArrayByParity * @Author: xiaof * @Description: 905. Sort Array By Parity * 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. * * 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. * * @Date: 2019/7/3 8:48 * @Version: 1.0 */ public class SortArrayByParity public int[] solution(int[] A) //先获取所有偶元素,然后获取所有奇元元素,还有一个方法就是交换前后位置 //吧所有偶数放到前面,奇数放后面,类似快排 int index1 = 0, index2 = A.length - 1; while(index1 < index2) //遍历起始第一个不是偶数的位置 while(index1 < A.length && (A[index1] & 1) == 0) index1++; //后面往前,找到第一个不是奇数 while(index2 > 0 && (A[index2] & 1) == 1) index2--; //交换位置 if(index1 < index2 && index1 < A.length && index2 > 0) //交换位置 int temp = A[index1]; A[index1] = A[index2]; A[index2] = temp; return A; public static void main(String args[]) int A[] = 3,1,2,4; SortArrayByParity fuc = new SortArrayByParity(); System.out.println(fuc.solution(A));
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