时间复杂度O(n)。 classSolution {public:voidmoveZeroes(vector<int>&nums) {intn =nums.size();if(n <2)return;intind1 =0;intind2 =0;while(ind1 < n && ind2 <n) {while(ind1 < n && nums[ind1] !=0) ind1++; ind2= ind1 +1;w
classSolution{public:voidmoveZeroes(vector<int>& nums){autoe = nums.end();//保存初始状态下数组的尾迭代器for(autoit = nums.begin(); it != nums.end();){if(it != e){//如果还没到达末尾if(*it ==0){//判断是否为0it = nums.erase(it);//删除操作会更新当前的迭代器nums.push_back(0...
[LeetCode] 283. Move Zeroes ☆(移动0到最后) 描述 给定一个数组nums,写一个函数,将数组中所有的0挪到数组的末尾,维持其他所有非0元素的相对位置。 举例: nums = [0, 1, 0, 3, 12], 函数运行后结果为[1, 3, 12, 0, 0] 解析 快慢指针,慢指针指向第一个0,快指针指向第一个非0. 代码 publicst...
leetcode 73 Set Matrix Zeroes 详细解答 leetcode 73 Set Matrix Zeroes 详细解答 解法1 此题如果空间复杂度为O(MN),则题目简单,只需要将为0的下标用一个集合保存起来,然后再判断所遍历的数字是否在集合内。 但根据题目要求,最好不要用空间复杂度为O(MN)的方法来做 解法2 只申请两个set用以保存数值为0...
LeetCode之Move Zeroes 1、题目 Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. For example, given nums = [0, 1, 0, 3, 12], after calling ...
LeetCode283. Move Zeroes(移动零)JAVA实现 Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. Example: Note: You must do this in-place witho......
思路:设定一个临时变量k=0,遍历数组nums,将非零元素与之前的零元素进行交换,维护变量k的值。 动画如下: 代码如下: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 1// 原地(in place)解决该问题2// 时间复杂度: O(n)3// 空间复杂度: O(1)4class Solution{5public:6voidmoveZeroes(vector<int>&...
【Leet Code】283. Move Zeroes Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. Example: Input: [0,1,0,3,12] Output: [1,3,12,0,0] Note:...
【LeetCode & 剑指offer 刷题笔记】目录(持续更新中...)Move ZeroesGiven an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. Example: Input: [0,1,0,3,12] Output: [1,3,12,0,0] Note: You must do this ...
class Solution { public void moveZeroes(int[] nums) { int setIndex = 0; int checkIndex = 0; while (checkIndex<nums.length){ if (nums[checkIndex] != 0){ nums[setIndex] = nums[checkIndex]; setIndex++; } checkIndex++; } for (int i=setIndex;i<nums.length;i++){ nums[i] = ...