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: 代码语言:t...
solution=Solution2() print(solution.removeElements(nums1,2)) print(solution.removeElements(nums2,2)) print(solution.removeElements(nums3,2)) print(nums3) # 暴力法 class Solution: def removeElement1(self, nums: [int], val:int)->int: lens=len(nums) i=lens # 总共需要循环i次,即列表长度...
AI代码解释 classSolution{public:intremoveElement(vector<int>&nums,int val){int count=0;for(int i=0;i<nums.size();i++){if(nums[i]!=val)nums[count++]=nums[i];}returncount;}};
对于2.1提到的暴力算法,如果在Python中如果调用 pop 抽取函数,则省掉第一个循环,相当于一个 for 搞掂。 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+=1ret...
代码(Python3) class Solution: def removeElement(self, nums: List[int], val: int) -> int: # l 表示不等于 val 的数字个数,也是下一个可以放入数字的下标,初始化为 0 l: int = 0 # 遍历剩余所有的数字 for r in range(len(nums)): # 如果当前数字不等于 val ,则 nums[r] 不需要移除,放入...
The order of elements can be changed. It doesn't matter what you leave beyond the new length. 代码: oj测试通过 Runtime: 43 ms 1classSolution:2#@param A a list of integers3#@param elem an integer, value need to be removed4#@return an integer5defremoveElement(self, A, elem):6length...
[Leetcode][python]Remove Element/移除元素 题目大意 去掉数组中等于elem的元素,返回新的数组长度,数组中的元素不必保持原来的顺序。 解题思路 双指针 使用头尾指针,头指针碰到elem时,与尾指针指向的元素交换,将elem都换到数组的末尾去。 代码 判断与指定目标相同...
Leetcode 27:Remove Element Leetcode 27:Remove Element Given an arraynumsand a valueval, remove all instances of that valuein-placeandreturn the new length. Do not allocate extra space for another array, you must do this by modifying the input arrayin-place with O(1)extra memory....
Leetcode 26 Remove Duplicates 题目描述 Remove Duplicates from Sorted Array Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. ...
Python源码: 1.解法一 fromtypingimportListclassSolution:defremoveElement(self,nums:List[int],val:int)->int:whilevalinnums:nums.pop(nums.index(val))returnlen(nums) 2.解法二 fromtypingimportListclassSolution:defremoveDuplicates(self,nums:List[int])->int: ...