[Leetcode 48] 18 4 Sum

Problem:

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)

Analysis:

It's very much like 3 Sum except that we need another variable start from the end of the given array

 

Code:

 1 class Solution {
 2 public:
 3     vector<vector<int> > fourSum(vector<int> &num, int target) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         vector<vector<int> > res;
 7         
 8         sort(num.begin(), num.end());
 9         
10         int a = 0;
11         while (a < num.size()) {
12             int b = num.size()-1;
13             while (b > a) {
14                 int c = a+1, d = b-1;
15                 
16                 while (c < d) {
17                     int sum = num[a] + num[b] + num[c] + num[d];
18                 
19                     if (sum == target) {
20                         vector<int> tmp;
21                         tmp.push_back(num[a]);
22                         tmp.push_back(num[c]);
23                         tmp.push_back(num[d]);
24                         tmp.push_back(num[b]);
25                     
26                         res.push_back(tmp);
27                     
28                         do {c++;} while (num[c]==num[c-1] && c < b);
29                         do {d--;} while (num[d]==num[d+1] && d > a);
30                     } else if ( sum < target) {
31                         c++;
32                     } else {
33                         d--;
34                     }
35                 }
36                 
37                 do {b--;} while (num[b] == num[b+1] && b > 0);
38             }
39             
40             do {a++;} while (num[a] == num[a-1] && a < num.size());
41         }
42         
43         return res;
44     }
45 };
View Code

 

 

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posted on 2013-05-25 10:42  freeneng  阅读(151)  评论(0)  编辑  收藏  举报

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