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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],
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) {
if(nums.size()<=2) return nums.size();
vector<int> num_new;
int len=1;
int n=1;
int begin=nums[0];
num_new.push_back(nums[0]);
for(int i=1;i<nums.size();i++)
{
if(nums[i]==begin)
{
if(n<2)
{
len++;
num_new.push_back(nums[i]);
}
n++;
}
else
{
begin=nums[i];
len++;
num_new.push_back(nums[i]);
n=1;
}
}
nums=num_new;
return len;
}
};
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size()<=2) return nums.size();
int len=0;
int n=0;
int begin=nums[0];
for(int i=1;i<nums.size();)
{
if(*(nums.begin()+i)==begin)
{
n++;
if(n>1)
{
nums.erase(nums.begin()+i);
}
else
i++;
}
else
{
begin=*(nums.begin()+i);
i++;
n=0;
}
}
return nums.size();
}
};class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size()<=2) return nums.size();
int rear=1;
for(int i=2;i<nums.size();i++)
{
if(!(nums[i]==nums[rear]&&nums[i]==nums[rear-1]))
nums[++rear]=nums[i];
}
return rear+1;
}
};Remove Duplicates from Sorted Array II
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原文地址：http://blog.csdn.net/sinat_24520925/article/details/46627937
