标签：blog io os ar for sp div on 问题
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;
}
标签：blog io os ar for sp div on 问题
原文地址：http://www.cnblogs.com/supernigel/p/4028708.html
