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.
典型的双指针问题，维护两个指针不断更新就可以了，代码如下：
1 class Solution { 2 public: 3 int removeDuplicates(vector<int>& nums) { 4 int sz = nums.size(); 5 if(!sz) return 0; 6 int start = 0;//第二个指针 7 int i = 0;//第一个指针 8 int count = 0; 9 int key = nums[0]; 10 for(; i < sz; ++i){ 11 if(nums[i] == key) 12 count++; 13 else{ 14 for(int k = 0; k < min(2, count); ++k) 15 nums[start++] = key;//更改数组元素，指针前移 16 key = nums[i]; 17 count = 1; 18 } 19 } 20 for(int k = 0; k < min(2, count); ++k) 21 nums[start++] = key; 22 return start; 23 } 24 };
LeetCode OJ：Remove Duplicates from Sorted Array II（移除数组中的重复元素II）
原文：http://www.cnblogs.com/-wang-cheng/p/4922080.html