4Sum

来源：Leetcode

原帖：http://oj.leetcode.com/problems/4sum/

题目：
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.
Note:
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)

代码：
1] no optimization, two pointers. O(n^3)

 class Solution {  
 public:  
   vector<vector<int> > fourSum(vector<int> &num, int target) {  
     int N = num.size();  
     vector<vector<int> > res;  
     if (N < 4) return res;  
     sort(num.begin(), num.end());  
     for (int i = 0; i < N - 3; ++i) {  
       if (i > 0 && num[i] == num[i - 1]) continue;   
       for (int j = i + 1; j < N - 2; ++j) {  
         if (j > i + 1 && num[j] == num[j - 1]) continue;   
         int twosum = target - num[i] - num[j];  
         int l = j + 1, r = N - 1;  
         while (l < r) {  
           if (l > j + 1 && num[l] == num[l - 1]) { // avoid duplicates  
             l++;   
             continue;   
           }   
           if (r < N - 1 && num[r] == num[r + 1]) { // avoid duplicates  
             r--;   
             continue;   
           }   
           int sum = num[l] + num[r];  
           if (sum == twosum) {  
             vector<int> four = {num[i], num[j], num[l], num[r]};  
             res.push_back(four);  
             l++; r--;  
           } else if (sum < twosum) {  
             l++;  
           } else {  
             r--;  
           }  
         }  
       }  
     }  
     return res;  
   }  
 };