Description
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
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
- 两个指针,一个指向原数组,一个指向新数组,然后一个计数器,在每一次相邻元素不同的时候更新
代码
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int len = nums.size();
if(len == 0) return 0;
int index = 0, count = 1;
for(int i = 1; i < len; ++i){
if(nums[i] != nums[i - 1]){
count = 1;
index++;
nums[index] = nums[i]; //直接覆盖,不要交换,交换会有bug
}
else{
if(count < 2){
index++;
count++;
nums[index] = nums[i];
}
}
}
return index + 1;
}
};