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)//once find the non-val number, swap non-val number with res index9nums[res++] =nums[i]...
Given a sorted array, remove the duplicates in place such that each element appear onlyonceand return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. For example, Given input array A =[1,1,2], Your function should return le...
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. 思路: 用count记录要删除元素数目。常用操作要熟记。 算法: public int removeElement(int[] nums,...
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(...
自己重写code, 有bug。要向上面一样,i从后往前扫,j+1为结果。跟move zero一样,要从后面那么i,j就都从后面,否则就都从前面 class Solution(object): def removeElement(self, nums, val): """ :type nums: List[int] :type val: int :rtype: int ...
Can you solve this real interview question? Remove Duplicates from Sorted Array - Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place [https://en.wikipedia.org/wiki/In-place_algorithm] such that each unique element
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
1909. 删除一个元素使数组严格递增 - 给你一个下标从 0 开始的整数数组 nums ,如果 恰好 删除 一个 元素后,数组 严格递增 ,那么请你返回 true ,否则返回 false 。如果数组本身已经是严格递增的,请你也返回 true 。 数组 nums 是 严格递增 的定义为:对于任意下标的 1
最后返回该变量值即可 java解法2: classSolution{publicintremoveElement(int[]nums,intval){//遍历数组,用数组最后元素替换相同元素,并缩短数组长度intn=nums.length;inti=0;while(i<n){if(nums[i]==val){nums[i]=nums[n-1];n--;}else{i++;}}returnn;}}...
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++){ ...