No.80 Remove Duplicates from Sorted Array ||

Remove Duplicates from Sorted Array|| 

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.

Tags： Array Two Pointers

移除有序数组中出现次数超过两次的数字
要求：原地移除，不能使用额外空间；返回新数组的长度

#include "stdafx.h"
#include <map>
#include <vector>
#include <iostream>
using namespace std;

class Solution
{
public:
    int removeDuplicates(vector<int> &nums)
    {//移除有序数组中出现次数超过两次的数字
     //要求：原地移除，不能使用额外空间；返回新数组的长度

        int size = nums.size();
        if(size <= 2)
            return size;
        int index = 0;//已确定的最后索引
        bool isTwice = false;//是否一出现两次
        for(int i=1; i<size; i++)
        {
            if(nums[i] != nums[index])
            {
                nums[++index] = nums[i];
                isTwice = false;
            }
            else
            {
                if(!isTwice)//重复，但此时为第二次出现
                {
                    nums[++index] = nums[i];
                    isTwice = true;
                }
            }
        }
        nums.erase(nums.begin()+index+1, nums.end());
        return index+1;
    }
};

int main()
{
    Solution sol;
    int data[] = {1,1,1,2,2,3,3,3,3,3,4,5};
    vector<int> test(data,data+sizeof(data)/sizeof(int));
    for(const auto &i : test)
        cout << i << " ";
    cout << endl;
    cout<< boolalpha << sol.removeDuplicates(test)<<endl;
        for(const auto &i : test)
        cout << i << " ";
    cout << endl;    
}

　　其实，还有一种更简单的做法，不用isTwice标识，直接比较nums[index-1]即可[因为是排序过的]!!

class Solution
{
public:
    int removeDuplicates(vector<int> &nums)
    {//移除有序数组中出现次数超过两次的数字
     //要求：原地移除，不能使用额外空间；返回新数组的长度

        int size = nums.size();
        if(size <= 2)
            return size;
        int index = 1;//已确定的最后索引

        for(int i=2; i<size; i++)//i从2开始
        {
            if(nums[i] != nums[index-1])
                nums[++index] = nums[i];
        }
        nums.erase(nums.begin()+index+1, nums.end());
        return index+1;
    }
};