Remove Duplicates from Sorted Array II
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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],
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.
思路:
题目的要求就是说数组中重复的元素最多只有两个。按照之前的移除算法, 当nums[i]!=nums[i-1]的时候说明前后两个元素不一致,则直接放到nums[index++] 中。当nums[i]==nums[i-1] 的时候只有找到这个元素的最后一个重复的元素的位置,将其放在nums[index++]中就可以了,因为之前肯定已经加入了一个,只要再放一个就行。
class Solution {public: int removeDuplicates(vector<int>& nums) { //主要是有序的条件,则这个时候重复的元素是在一起的 if(nums.size()<2) return nums.size(); int index=1; for(int i=1;i<nums.size();i++) { if(nums[i]!=nums[i-1]) //不重复元素 nums[index++]=nums[i]; else { while(i<nums.size()&&nums[i]==nums[i-1]) ++i; nums[index++]=nums[--i]; } } nums.resize(index); return index; }};
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