只有到0后面的非零元素,才会和0进行交换,最终将全部的0移到列表右边。 class Solution: def moveZeroes(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ prev = 0 ## 从第一个位置开始,所以初始化=0 for i in range(len(nums)): if nums[i...
1: class Solution { 2: public: 3: voidmoveZeroes(vector<int>& nums) { 4:for(int zero_index =0, none_zero_index =0; 5: none_zero_index < nums.size() && zero_index < nums.size(); 6: ) { 7:if(nums[zero_index] !=0) { 8: zero_index++; 9: none_zero_index = zero_ind...
为此,需要记录初始状态下数组的尾迭代器,然后在循环体里判断迭代器是否已到了这个尾迭代器,如果到了,直接break。 于是就有了下面的代码: 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(...
//时间复杂度: O(n)//空间复杂度: O(n)classSolution{public:voidmoveZeroes(vector<int>&nums){vector<int>nonZeroElements;// 将vec中所有非0元素放入nonZeroElements中for(inti=0;i<nums.size();i++)if(nums[i])nonZeroElements.push_back(nums[i]);// 将nonZeroElements中的所有元素依次放入到nums...
leetcode上第283号问题:Move Zeros 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,⽽维持其他所有非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] 解法一 思路:创建一个临时数组nonZeroElements,遍历nums,将nums中非0元素赋值到nonZeroEleme...
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]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: Note: You must do this in-place witho...猜你喜欢LeetCode.283. Move Zeroes https://leetcode.com/problems/...
Hint Given an integer arraynums, move all0's to the end of it while maintaining the relative order of the non-zero elements. Notethat you must do this in-place without making a copy of the array. Example 1: Input:nums = [0,1,0,3,12]Output:[1,3,12,0,0] ...
【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:...
代码 public class Solution { public void moveZeroes(int[] nums) { int cnt = 0, pos = 0; // 将非0数字都尽可能向前排 for(int i = 0; i < nums.length; i++){ if(nums[i] != 0){ nums[pos]= nums[i]; pos++; } }