因为要求是返回修改后的长度并只考虑该长度的数组,那么就不用考虑该长度之后的数组,所以只需得到索引 j 的值,不用再把索引 j 的值改为索引 i的值。 Java: 代码语言:txt 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...
java代码如下: 1publicclassSolution {2publicintremoveElement(int[] A,intelem) {3intputhere = 0;4intn =A.length;5for(inti = 0;i < A.length;i++){6if(A[i] !=elem){7A[puthere] =A[i];8puthere++;9}10else{11n--;12}13}14returnn;15}16}...
res记录了所有不需要删除的数字。 Java Solution: Runtime beats 95.05% 完成日期:03/27/2017 关键词:Array 关键点:Two Pointers 1publicclassSolution2{3publicintremoveElement(int[] nums,intval)4{5intres = 0;//count non-val numbers67for(inti=0; i<nums.length; i++)8if(nums[i] != val)//on...
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(...
【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之Remove Element 1、题目 Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. The order of elements can be changed. It doesn't matter ...
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...
java解法1: 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...
用一个指针记录不含给定数字的数组边界,另一个指针记录当前遍历到的数组位置。只有不等于给定数字的数,才会被拷贝到子数组的边界上。 代码 public class Solution { public int removeElement(int[] nums, int val) { int pos = 0; for(int i = 0; i < nums.length; i++){ ...