Remove Duplicates from Sorted ListII

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].

原始思路:

找一个标签数字，遍历向量元素，如果相等，则加1，如果大于2,则continue遍历，如果不同标签数字换成新的数字

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{vector<int> a = { 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4,4 };int k = 0;int key = a[0];int index = 0;int i = 0;while (i < a.size()){if (a[i] == key){k++;if (k<3)a[index++] = key;else {i++;continue;}}else{key = a[i];a[index++] = key;k = 1;}i++;}for (int i = 0; i < a.size(); i++)cout << a[i] << endl;system("pause");return 0;
}

应为允许一个重复，所以判断就不是相邻而是相隔一个的数字是否不一样。

不一样就需要更新值，但注意到这题更新数组不能是当前的状态，而需要是更新前一个不同的数，用temp存当前值以便于下一次赋值。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{vector<int> a = { 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4,4 };int k = 0;/*int key = a[0];int index = 0;int i = 0;while (i < a.size()){if (a[i] == key){k++;if (k<3)a[index++] = key;else {i++;continue;}}else{key = a[i];a[index++] = key;k = 1;}i++;}*/int num = 1, i, temp = a[1];for (i = 2; i<a.size(); ++i)if (a[i] != a[i - 2]){a[num++] = temp;temp = a[i];}a[num++] = temp;for (int i = 0; i < a.size(); i++)cout << a[i] << endl;system("pause");return 0;
}

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