leetcode 905. 按奇偶排序数组(Sort Array By Parity)

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给定一个非负整数数组 A，返回一个由 A 的所有偶数元素组成的数组，后面跟 A 的所有奇数元素。

你可以返回满足此条件的任何数组作为答案。

示例：

输入：[3,1,2,4]
输出：[2,4,3,1]
输出 [4,2,3,1]，[2,4,1,3] 和 [4,2,1,3] 也会被接受。

提示：

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

原题链接

实现方式：双指针，非原地自排，时间和空间O(n)

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int[] ans = new int[A.length];
        int l=0,r=A.length-1;
        for(int i=0;i<A.length;i++){
        	if(A[i]%2==0)
        		ans[l++]=A[i];
        	else
        		ans[r--]=A[i];
        }
        return ans;
    }
}

下面再给4个优质代码，看到过瘾吧~~  From Beat Top 4

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int length = A.length;
        int B[] = new int[length];
        int j = 0;
        int k = 0;
        for (int i = 0; i < length; i++) {
            if (A[i] % 2 == 0) {
                B[j] = A[i];
                j++;
            } else {
                B[length - k - 1] = A[i];
                k++;
            }
        }
        
        return B;
    }
    
    public static void main(String[] args) {
        int A[] = {3,1,2,4};
        Solution solution = new Solution();
        int B[] = solution.sortArrayByParity(A);
        for (int i = 0; i < B.length; i++) {
            System.out.print(B[i]);
        }
    }
}

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int index = 0;
        int[] res = new int[A.length];
        for (int i = 0; i < A.length; i++) {
            if (A[i] % 2 == 0) {
                res[index] = A[i];
                index ++;
            }
        }

        for (int i = 0; i < A.length; i++) {
            if (A[i] % 2 != 0) {
                res[index] = A[i];
                index ++;
            }
        }

        return res;
    }
}

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int[] a = new int[A.length];
        int b = 0;
        for (int i = 0; i < A.length; ++i) {
        	if ((A[i] & 1) != 1) a[b++] = A[i];
        }
        for (int i = 0; i < A.length; ++i) {
        	if ((A[i] & 1) == 1) a[b++] = A[i];
        }
        return a;
    }
}

class Solution {
    public int[] sortArrayByParity(int[] A) {
         int[] a = new int[A.length];
        int b = 0;
        for (int i = 0; i < A.length; ++i) {
        	if ((A[i] & 1) != 1) a[b++] = A[i];
        }
        for (int i = 0; i < A.length; ++i) {
        	if ((A[i] & 1) == 1) a[b++] = A[i];
        }
        return a;
    }
}