LeetCode OJ 80. Remove Duplicates from Sorted Array II

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题目

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 matter what you leave beyond the new length.

解答

这题又是一遍AC。。。只能说LeetCode上水题有点多，这居然都是Medium难度。。。

拿一个变量计数一下重复次数就可以了，每次遇到新的数就将这个变量赋值为1，如果这个变量达到2且遇到和之前重复的数，就移除。

下面是AC的代码：

class Solution {public: int removeDuplicates(vector<int>& nums) { if(nums.size() == 0){ return 0; } int last = nums[0]; int count = 1; int dup = 1; for(vector<int>::iterator iter = nums.begin() + 1; iter != nums.end(); iter++){ if(*iter == last){ if(dup < 2){ dup++; count++; } else{ nums.erase(iter); iter--; } } else{ dup = 1; count++; last = *iter; } } return count; }};

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