【LEETCODE】41、905. Sort Array By Parity

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));
    }
}

原文地址：https://www.cnblogs.com/cutter-point/p/11128151.html