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)
classSolution{publicvoidmoveZeroes(int[] nums){intp=;for(inti=; i < nums.length; i++){if(nums[i]!=){inttemp= nums[p]; nums[p++]= nums[i]; nums[i]= temp;}}} 4.3> 实现3：非0数字左移，剩余置为0 classSolution{publicvoidmoveZeroes(int[] nums){intp=;for(inti=; i <...
解法一:就冒泡排序的思想 冒泡排序算法:王几行xing:【Python入门算法6】冒泡排序 Bubble Sort 的三种实现方法 我们先试试基于冒泡排序进行改造,将0元素逐一换到右边一位,一直冒到最右边。 class Solution: def moveZeroes(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-plac...
Minimize the total number of operations 限制不能用array的拷贝,所以想法就是遍历遇到非零的元素,判断非零元素下一个元素是否为0,是0那么就swap,非零就index++ classSolution {public:voidmoveZeroes(vector<int>&nums) {intindex =0;for(inti =0; i < nums.size(); i++){if(nums[i] !=0){if(nums...
AI代码解释 classSolution{publicvoidmoveZeroes(int[]nums){int[]zeros=newint[nums.length];Arrays.fill(zeros,-1);int index=0,p=0;for(int i=0;i<nums.length;i++){if(nums[i]==0)zeros[index++]=i;elseif(zeros[p]!=-1&&zeros[p]<i){nums[zeros[p++]]=nums[i];nums[i]=0;zeros[in...
维护两个指针,不断的往前移动。 AC代码: publicclassSolution {publicvoidmoveZeroes(int[] nums) {inti=0,j=0;while(i<nums.length){if(nums[i]==0){ j=Math.max(i,j);while(j<nums.length && nums[j]==0) j++;if(j==nums.length)return; ...
LeetCode "Move Zeroes" Solution {public:moveZeroes(vector<int>&nums) { size_t n=nums.size();if(n <2)return;inti =0, j =0;while(j <n) {if(nums[i] !=0) { i++; j = i +1; }elseif(nums[j] ==0) { j++; }elseif(j <n)...
classSolution{publicvoidmoveZeroes(int[]nums){int n=nums.length,left=0,right=0;while(right<n){if(nums[right]!=0){swap(nums,left,right);left++;}right++;}}publicvoidswap(int[]nums,int left,int right){int temp=nums[left];nums[left]=nums[right];nums[right]=temp;}} ...
class Solution { public: void moveZeroes(vector<int>& nums) { int k = 0; //当前可以放的位置 for (int i = 0; i < nums.size(); i++) { //i表示当前遍历到的位置,红色指针 if (nums[i]) nums[k++] = nums[i]; } //当i越界之后,k表示当前可以放的最后一个位置,但是 ...
class Solution {public void moveZeroes(int[] nums) {if (nums == null || nums.length == 0)return;int cur = 0;for (int i = 0; i < nums.length; i++) {if (nums[i] != 0)nums[cur++] = nums[i];}while (cur < nums.length) {nums[cur++] = 0;}}} ...