public void swap(int [] Array,int i,int j){ int temp=Array[i]; Array[i]=Array[j]; Array[j]=temp; }
此解法的时间复杂度是O(N),空间复杂度是O(1)。 publicint[] sortArrayByParityII4(int[] A) {inti=0, j = A.length-1, n = A.length;while(i < n && j >=1) {if(A[i]%2==1&& A[j]%2==0) {inttem=A[j]; A[j] = A[i]; A[i] = tem; }if(A[i]%2==0) { i +=2...
leetcode 922. Sort Array By Parity II ... LeetCode #922. 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; an......
package leetcode func sortArrayByParityII(A []int) []int { if len(A) == 0 || len(A)%2 != 0 { return []int{} } res := make([]int, len(A)) oddIndex := 1 evenIndex := 0 for i := 0; i < len(A); i++ { if A[i]%2 == 0 { res[evenIndex] = A[i] even...
LeetCode-922. Sort Array By Parity II i++文章分类运维 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; ...
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LeetCode-Sort Array By Parity Description: 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....
/** * 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 ...
这类问题不能用上述排序方法解决,更多是考量对数组元素排布的处理逻辑。 相关LeetCode题: 75. Sort Colors 题解 922. Sort Array By Parity II 题解 280. Wiggle Sort 题解 324. Wiggle Sort II 题解 1054. Distant Barcodes 题解 767. Reorganize String 题解 969. Pancake Sorting 题解编辑...
Hint 1 Try to separate the elements at odd indices from the elements at even indices. Hint 2 Sort the two groups of elements individually. Hint 3 Combine them to form the resultant array. Similar Questions Sort Array By Parity Easy Sort Array By Parity II Easy Discussion (19) ...