Remove Duplicates from Sorted Array II

Remove Duplicates from Sorted Array II

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.

discuass的讲解：

https://discuss.leetcode.com/topic/7673/share-my-o-n-time-and-o-1-solution-when-duplicates-are-allowed-at-most-k-times

https://discuss.leetcode.com/topic/17180/3-6-easy-lines-c-java-python-ruby

提示上采用双指针。

class Solution { int remove(vector & nums, int k) { int n=nums.size(); int i=1,j=1,cnt=1; //i指向原来的数组,j指向新的数组 while(i & nums) { if(nums.size()==0) return 0; return remove(nums,2); } }; 更简洁的：判断nums[i]>nums[i-2]?与否

int removeDuplicates(vector & nums) { int i = 0; for (int n : nums) if (i < 2 || n > nums[i-2]) nums[i++] = n; return i; }