Given two sorted integer arrays A and B, merge B into A as one sorted array. Note: You may assume that A has enough space (size that is greater or equal tom+n) to hold additional elements from B. The number of elements initialized in A and B aremandnrespectively. 代码:oj测试通过 R...
【leetcode python】 88. Merge Sorted Array #-*- coding: UTF-8 -*- class Solution(object): def merge(self, nums1, m, nums2, n): """ :type nums1: List[int] :type m: int :type nums2: List[int] :type n: int :rtype: void Do not return anything, modify nums1 in-place ...
最后只有while(n > 0),因为如果保留的是m >0,就保持原来的数组就好了 class Solution { public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { if(nums1.empty() || nums2.empty()) return; while(m > 0 && n > 0){ if(nums1[m-1] < nums2[n-1]){ nums1...
Merge nums1 and nums2 into a single array sorted in non-decreasing order. The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that...
Javapublic class Solution { // 归并排序 /** * 列表大小等于或小于该大小,将优先于 mergeSort 使用插入排序 */ private static final int INSERTION_SORT_THRESHOLD = 7; public int[] sortArray(int[] nums) { int len = nums.length; ...
Merge Sorted Array 这个题目的意思是把两个排好序的序列合并为一个排好序的序列,本来我最开始想的是直接把两个序列拼接在一起,然后用python列表里面自带的sort函数排个序就Ok了,但是这样做的复杂度至少是O((m+n)log(m+n))的,另外一种O(m+n)做法是从后面往前面排,比如合并后的最后一个数A[m+n-1]就...
LeetCode 0088. Merge Sorted Array合并两个有序数组【Easy】【Python】【双指针】 题目 英文题目链接 Given two sorted integer arraysnums1andnums2, mergenums2intonums1as one sorted array. Note: The number of elements initialized innums1andnums2aremandnrespectively. ...
void merge(int A[], int m, int B[], int n) { int k = m + n; while (k-- > 0) A[k] = (n == 0 || (m > 0 && A[m-1] > B[n-1])) ? A[--m] : B[--n]; } }; 可读性较好: class Solution { public: ...
链接:https://leetcode-cn.com/problems/squares-of-a-sorted-array python # 有序数组的平方 class Solution: def sortedSquares(self, nums: [int]) -> [int]: """ 双指针,时间O(n),空间除存储空间外O(1) 思路: 1.i,j即左右指针,每次移动,逆序将对应大的平方加入新开辟的数组, ...
代码(Python3) classSolution:defmerge(self,nums1:List[int],m:int,nums2:List[int],n:int)->None:"""Do not return anything, modify nums1 in-place instead."""# i/j 分别表示 nums1/nums2 中还未使用的最大数的下标i,j=m-1,n-1# k 表示 nums1 中下一个该填充的位置。k:int=m+n-1...