Array篇easy难度之奇偶分区

关键词

奇偶分区 two pointers,

题目描述

https://leetcode.com/problems/sort-array-by-parity-ii

Given an array A of non-negative integers, half of the integers in A are odd, and 
half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

 

Example 1:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
 

Note:

2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000

博主提交的代码

IN-PLACE 版本

思路错了，错误代码，别看,这个问题本质就是需要同时判断奇偶再进行交换，我这个思路已经不进行

class Solution {
    public int[] sortArrayByParityII(int[] A) {
        int evenIndex = 0;
        int oddIndex = 1;
        for(int i = 0; i < A.length;){
            if(isEven(i) && !isEven(A[i])){
                oddIndex+=2;
            } else if(!isEven(i) && isEven(A[i])){
                int tmp = A[evenIndex];
                A[i]  = tmp;
                oddIndex+=2;
            } else if(isEven(i)) {
                evenIndex+=2;
                i++;
            } else if(!isEven(i)){
                oddIndex+=2;
                i++;
            }
        }
        return A;
        
    }
    public boolean isEven(int input){
        if( (input & 1) == 0){
            return true;
        } else{
            return false;
        }
    }
}

非IN-PLACE 版本

class Solution {
    public int[] sortArrayByParityII(int[] A) {
        int i = 0, j = 1, n = A.length;
        while (i < n && j < n) {
            while (i < n && A[i] % 2 == 0) {
                i += 2;
            }
            while (j < n && A[j] % 2 == 1) {
                j += 2;
            }
            if (i < n && j < n) {
                swap(A, i, j);
            }
        }
        return A;
    }
    private void swap(int[] A, int i, int j) {
        int temp = A[i];
        A[i] = A[j];
        A[j] = temp;
    }
}

其他人优秀的解法

https://leetcode.com/problems/sort-array-by-parity-ii/discuss/181160/Java-two-pointer-one-pass-inplace
INPLACE

class Solution {
    public int[] sortArrayByParityII(int[] A) {
        int i = 0, j = 1, n = A.length;
        while (i < n && j < n) {
            while (i < n && A[i] % 2 == 0) {
                i += 2;
            }
            while (j < n && A[j] % 2 == 1) {
                j += 2;
            }
            if (i < n && j < n) {
                swap(A, i, j);
            }
        }
        return A;
    }
    private void swap(int[] A, int i, int j) {
        int temp = A[i];
        A[i] = A[j];
        A[j] = temp;
    }
}