双指针法)classSolution{public:intremoveElement(vector<int>& nums,intval){intres =0;// 存储去重后数组元素个数for(inti =0; i < nums.size(); ++i)// 快指针i遍历数组if(nums[i] != val)// 若快指针指向元素不是目标元素nums[res++] = nums[i];// 则慢指针res+1,且慢指针指向快指针当前元素...
AI代码解释 class Solution { public int removeElement(int[] nums, int val) { int i=0,j=nums.length-1;//i-左指针;j-右指针 while (i<=j){ if(nums[i]==val){ nums[i]=nums[j];//得到索引j的值,无需把索引j的值改为索引i的值 j--; }else i++; } return j+1; } } Python3:...
1classSolution {2public:3intremoveElement(vector<int>& nums,intval) {4inti=0;5intp=0;6intlength=nums.size();7for(i;i<length;i++){8if(nums[i]!=val)9nums[p++]=nums[i];10}11returnp;12}13};
classSolution:defremoveElement(self,nums:List[int],val:int)->int:i=0##从第一个元素开始whilei<len(nums):##遍历每一个元素ifnums[i]==val:nums.pop(i)##删除目标元素else:##继续前进i+=1returnnums,i##删掉最后一个的时候,i的取值就是非valnums的长度##顺便也把nums返回,查看效果 上面的 pop(...
27 | LeetCode | Remove Element 题目: Given an array and a value, remove all instances of that value in place and return the new length. The order of elements can be changed. It doesn't matter what you leave beyond the new length....
【Leetcode】Remove Element 题目链接: 题目: Given an array and a value, remove all instances of that value in place and return the new length. The order of elements can be changed. It doesn't matter what you leave beyond the new length....
LeetCode Remove Element删除元素 1 class Solution { 2 public: 3 int removeElement(int A[], int n, int elem) { 4 int *p=A,*e=&A[n-1]; 5 int i,num=n; 6 for(i=0;i<n;i++){ //一共要对比n次,不能用n来处理,会影响循环...
Can you solve this real interview question? Remove Linked List Elements - Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head. Example 1: [https://assets.leetc
LeetCode——Remove Element Given an array and a value, remove all instances of that value in place and return the new length. 52720 Remove Element 问题:删除数组中和elem相等的元素,并且返回新数组大小。英语不好。。。读错题了。。 class Solution { public: int remo... 2K80 HashMap remove Conc...
classSolution{publicintremoveElement(int[]nums,intval){//两个指针intidx=-1;for(inti=0;i<nums.length;i++){if(nums[i]!=val){nums[++idx]=nums[i];}}returnidx+1;}} 思路2:看一个例子,nums = [4,1,2,3,5], val = 4。在这个例子中,如果使用解法1的思路,会产生很多不必要的复制过程。