【Leetcode】Remove Duplicates from Sorted Array II

题目：对上一题的延伸，每个数字可以出去2次。

思路：还是设置两个下标。第一个lenxb标记已去重的地方，第二个i标记待处理的位置。每次比较时，比较lenxb和lenxb-1两个位置，如果都相等，说明出现超过两次了；否则满足要求。

注意：通过上面的思路可知，特判情况是长度小于等于2时。

PS:提交后16ms，在此题的提交时间第一队列最前位置。Yes！

代码：

class Solution {
public:int removeDuplicates(vector<int>& nums) {if(nums.size()<=2) return nums.size();int lenxb=1;//长度的下标for(int i=2;i<nums.size();++i){if(nums[i]==nums[lenxb] && nums[i]==nums[lenxb-1]) continue;nums[++lenxb]=nums[i];}return lenxb+1;}
};

转载于:https://www.cnblogs.com/buxizhizhou/p/4684490.html

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