classSolution {publicint[] sortArrayByParity(int[] A) {if(A ==null)returnnull;int[] ret =newint[A.length];intk = A.length-1, j=0;for(inti=0; i<A.length; i++) {if(isEven(A[i])) { ret[j++] =A[i]; }else{ ret[k--] =A[i]; } }returnret; }privatebooleanisEven(in...
public void swap(int [] Array,int i,int j){ int temp=Array[i]; Array[i]=Array[j]; Array[j]=temp; }
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; an...Leetcode学习笔记:#922. Sort Array By Parity II Leetcode学习笔记:#922. Sort Array By Parity II Give...
解法:我们可以定义两个变量st和ed,st用于指示从首部开始的遇到的奇数,ed用于指示从尾部开始遇到的偶数,将这两个位置的元素进行交换,一直重复这个操作知道st > ed,这样所有的偶数就被交换到了首部,奇数被交换到了尾部; class Solution { public int[] sortArrayByParity(int[] A) { int st = 0; int ed = ...
922. Sort Array By Parity II # 题目# 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...
classSolution{ public: vector<int>sortArrayByParityII(vector<int>&A) { intn=A.size(); vector<int>odd,even; for(inti=0;i<n;i++) { if(A[i]%2==0) { even.push_back(A[i]); } else{ odd.push_back(A[i]); } } A.clear(); ...
/** * 922. Sort Array By Parity II *https://leetcode.com/problems/sort-array-by-parity-ii/description/ * * 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 ...
:pencil2: 算法相关知识储备 LeetCode with Python :books:. Contribute to HuberTRoy/leetCode development by creating an account on GitHub.
:pencil2: 算法相关知识储备 LeetCode with Python :books:. Contribute to lukelucode/leetCode development by creating an account on GitHub.
*/int*sortArrayByParity(int*A,intASize,int*returnSize){inteven=0,odd=ASize-1;while(even<odd){if(A[even]%2!=0){if(A[odd]%2==0){inttmp=A[odd];A[odd]=A[even];A[even]=tmp;even++;}odd--;}else{even++;}}*returnSize=ASize;returnA;}...