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.
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int n = nums.size(), skip = 0;
if(n <= 2) return n;
int cnt = 0, index = 0;
for(int i = 0; i<n; i++){
if(i > 0 && nums[i] == nums[i-1]){
cnt++;
if(cnt >= 3)
continue;
}
else cnt = 1;
nums[index++] = nums[i];
}
return index;
}
};
用了index指明了结果数组中每个地方的正确数字，不必每次遇到重复数字时全局移动几乎整个后面的数组，用cnt变量表明到底出现了几次。
