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4Sum.cpp
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4Sum.cpp
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/*
Author: Weixian Zhou, ideazwx@gmail.com
Date: June 21, 2012
Problem: 4 Sum
Difficulty: medium
Source: http://www.leetcode.com/onlinejudge
Notes:
Given an array S of n integers, are there elements a, b, c, and d in S such that
a + b + c + d = target? Find all unique quadruplets in the array which gives the sum
of target.
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
A solution set is:
(-1, 0, 0, 1)
(-2, -1, 1, 2)
(-2, 0, 0, 2)
Solution:
The idea is different with 3 Sum problem and 3 Sum Closest problem. Frist compute the sum
of every two elements and sort them, this takes O(n^2) + O(n*lgn). Then for every element
in the sorted array, binary search for the corresponding element that give the sum equals
target, this takes O(n^2 * lgn). Thus the total time complexity is O(n^2*lgn)
A good STL sort comparison function guide: http://www.codeproject.com/Articles/38381/STL-Sort-Comparison-Function
This code doesn't pass the large judege of Leetcode, but when I test the wrong test cases
on my own machine, the code runs the correct output. I don't why.
*/
#include <vector>
#include <set>
#include <climits>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
class Triple {
public:
int sum;
int val1;
int val2;
Triple (int x) {
this->sum = x;
}
Triple (int x, int y, int z) {
this->sum = x;
this->val1 = y;
this->val2 = z;
}
bool operator < (const Triple& rhs) const {
return this->sum < rhs.sum;
}
};
class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
vector<Triple> sum2;
vector<Triple>::iterator it;
set<vector<int> > solset;
vector<int> quar(4);
vector<vector<int> > sol;
if (num.size() < 4) {
return sol;
}
for (int i = 0; i < num.size() - 1; i++) {
for (int j = i + 1; j < num.size(); j++) {
sum2.push_back(Triple(num[i] + num[j], i, j));
}
}
sort(sum2.begin(), sum2.end());
for (int i = 0; i < sum2.size(); i++) {
it = lower_bound(sum2.begin(), sum2.end(), Triple(target - sum2[i].sum));
Triple t1 = sum2[i];
Triple t2 = *it;
if (t1.val1 != t2.val1 && t1.val1 != t2.val2
&& t1.val2 != t2.val1 && t1.val2 != t2.val2
&& t1.sum + t2.sum == target) {
quar[0] = num[t1.val1];
quar[1] = num[t1.val2];
quar[2] = num[t2.val1];
quar[3] = num[t2.val2];
sort(quar.begin(), quar.end());
solset.insert(quar);
}
}
sol.resize(solset.size());
copy(solset.begin(), solset.end(), sol.begin());
return sol;
}
};