[LeetCode]3 Sum

思路: 1.将数组排序，

          2.a 遍历 数组a[0]....a[n-1];         

          3.当 a=a[i]  时   后面的问题 就是 :  a[i+1] 到 a[n-1]中  b+c =-a  （编程之美 2.12 快速寻找满足条件的两个数  ）      

                       记 b=a[j]=a[i-1]     c=a[k]=a[n-1]   

   若 b+c  < -a ，j++; 

b+c > -a  ，j--;    

b+c=-a 记录下来，并j++;

  4.还有一个问题 就是unique triplet,   所以 a=a[i] 要判断是否和a[i-1]相等，若相等，子问题已经解答。

                                                                              也要判断 b和c  是否和之前的相同，若相同，就已经判断过了。

// 3Sum.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
bool myComp(int x1, int x2)
{
	return x1 < x2;
}
class Solution {
public:
	vector<vector<int> > threeSum(vector<int> &num) {
		sort(num.begin(), num.end(), myComp);
		int p1 = 0, p2 = 0;
		vector<vector<int> > res;
		if (num.size() < 3)
			return res;
		for (int i = 0; i < num.size()-2; i++)
		{
			if (i>0 && num[i] == num[i - 1])continue;
			int sum = -num[i];
			p1 = i + 1;
			p2 = num.size() - 1;
			while (p1<p2)
			{
				//while (p1!=1&&num[p1] == num[p1 -1])p1++;
				while ((p1!=i+1)&&num[p1] == num[p1-1])p1++;
				while ((p2!=num.size()-1)&&(p2>0)&&(num[p2] == num[p2 +1]))p2--;
				if (p1 >= p2)break;
				//cout << "(" << i << "," << p1 << "," << p2 << ")" << endl;
				if (num[p1] + num[p2] == sum)
				{
					vector<int>tempV;
					tempV.push_back(num[i]);
					tempV.push_back(num[p1]);
					tempV.push_back(num[p2]);
					res.push_back(tempV);
					//cout << "**" << i << "," << p1 << "," << p2 << endl;
					p2--;
				}
				else if (num[p1] + num[p2] > sum)
				{
					p2--;
				}
				else if (num[p1] + num[p2] < sum)
				{
					p1++;
				}
				
			}
			
		}
		return res;
	}
};
int _tmain(int argc, _TCHAR* argv[])
{
	vector<int>test;
	/*
	test.push_back(-1);
	test.push_back(0);
	test.push_back(1);
	test.push_back(2);
	test.push_back(-1);
	test.push_back(-4);
	*/
	
	test.push_back(0);
	test.push_back(0);
	test.push_back(0);
	test.push_back(0);
	test.push_back(0);
	test.push_back(0);
	Solution ss;
	vector<vector<int> > res = ss.threeSum(test);
	//cout << res.size()<<endl;
	for (int i = 0; i < res.size(); i++)
	{
		for (int j = 0; j < res[i].size(); j++)
			cout << res[i][j] << ",";
		cout << endl;
	}
	system("pause");
	return 0;
}