标签：
Problems:
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.
解法一：在Remove Duplicates from sorted Array的基础上加入计数count用来判断重复的次数
1 class Solution { 2 public: 3 int removeDuplicates(vector<int>& nums) { 4 if(nums.size()==0) return 0; 5 6 int index=0; 7 int count=1; 8 for(int i=1;i<nums.size();i++) 9 { 10 if(nums[index]!=nums[i]) 11 { 12 nums[++index]=nums[i]; 13 count=1; 14 } 15 else if(nums[index]==nums[i]&&count<=1) 16 { 17 nums[++index]=nums[i]; 18 count++; 19 } 20 21 } 22 return index+1; 23 24 } 25 };
解法二：
代码简洁聪明
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size()<=2) return n;
int index=2;
for(int i=2;i<nums.size();i++)
{
if(nums[i]!=nums[index-2])
nums[index++]=nums[i];
}
return index;
}
};
LeetCode:Remove Duplicates from Sorted Array II
标签：
原文地址：http://www.cnblogs.com/xiaoying1245970347/p/4551109.html
