LeetCode Online Judge 题目C# 练习 - 4SUM

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)

 1         public static List<List<int>> fourSum(int[] S, int target)

 2         {

 3             List<List<int>> ret = new List<List<int>>();

 4             if(S.Length < 4)

 5                 return ret;

 6 

 7             Array.Sort(S);

 8             string str;

 9             Dictionary<string, bool> map = new Dictionary<string,bool>();

10 

11             for(int a = 0; a <= S.Length - 4; a++)

12             {

13                 for(int b = a + 1; b <= S.Length - 3; b++)

14                 {

15                     int c = b+1;

16                     int d = S.Length - 1;

17                     while(c < d)

18                     {

19                         if(S[a] + S[b] + S[c] + S[d] == target)

20                         {

21                             str = S[a].ToString() + S[b].ToString() + S[b].ToString() + S[c].ToString();

22                             if(map.ContainsKey(str))

23                             {

24                                 c++;

25                                 d--;

26                             }

27                             else

28                             {

29                                 List<int> quadruplet = new List<int> { S[a], S[b], S[c], S[d]};

30                                 map.Add(str, true);

31                                 ret.Add(quadruplet);

32                                 c++;

33                                 d--;

34                             }

35                         }

36                         else if (S[a] + S[b] + S[c] + S[d] < target)

37                             c++;

38                         else

39                             d--;

40                     }

41                 }

42             }

43 

44             return ret;

45         }

代码分析:

　　这一题比前面3SUM和3SUM Cloeset多了一个指针，但是原理还是相同，先排列数组，然后BRUTE FROCE。这使整个算法的时间复杂度为O(n³), 不知道没有更好的算法，请不吝赐教。

　　同样的用了一个Dictionary 集合帮助做防止重复的验证。