LeetCode OJ 80. Remove Duplicates from Sorted Array II

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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