排序数组去重题,保留重复两个次数以内的元素,不申请新的空间. 解法一: 因为已经排好序,所以出现重复的话只能是连续着,所以利用个变量存储出现次数,借此判断. Runtime: 20 ms, faster than 19.12% of C++ online submissions for Remove Duplicates from Sorted Array II. class Solution{public:  int removeDuplicates(vector<int> &num…

LeetCode 80 Remove Duplicates from Sorted Array II [Array/auto] <c++> 给出排序好的一维数组,如果一个元素重复出现的次数大于两次,删除多余的复制,返回删除后数组长度,要求不另开内存空间. C++ 献上自己丑陋无比的代码.相当于自己实现一个带计数器的unique函数 class Solution { public: int removeDuplicates(std::vector<int>& nums) {…

https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/discuss/27976/3-6-easy-lines-C%2B%2B-Java-Python-Ruby 描述 Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array A = [1,1…

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra mem…

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra mem…

Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array nums = [1,1,1,2,2,3], Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't…

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra mem…

题目链接:https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/#/description 给定一个已经排好序的数组,数组中元素存在重复,如果允许一个元素最多可出现两次,求出剔除重复元素(出现次数是两次以上的)后的数组的长度. For example, Given sorted array nums = [1,1,1,2,2,3], Your function should return length = 5,…

原题地址 简单模拟题. 从先向后遍历,如果重复出现2次以上,就不移动,否则移动到前面去 代码: int removeDuplicates(int A[], int n) { ) return n; ; ; ; i < n; i++) { ]) { dupSum++; } else dupSum = ; ) { A[len] = A[i]; len++; } } return len; }…

和第一题不同的地方是,容忍两次重复 虽然题目上说只需要长度,但是否检测的时候如果数组不跟着改变也是不行的 没说清楚题意 自己是用双指针做的,看了大神的答案更简单 public int removeDuplicates(int[] nums) { int i = 0; for (int n : nums) if (i < 2 || n > nums[i-2]) nums[i++] = n; return i; }…