[LeetCode] 3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique
triplets in the array which gives the sum of zero.
Note:
• Elements in a triplet (a, b, c) must be in non-descending order. (ie, a b c)

• The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4}.
A solution set is:
(-1, 0, 1)
(-1, -1, 2)

分析：先排序，然后取出一个数后，在余下的数中找其他两个，即2sum，另外注意，由于题目中指定了unique，不能有重复，所以在b++，c--，a++时都要对重复情况做处理。

PS；其实直接使用系统的sort就行，我这里实现了一个mergeSort，算是练手吧。。

PSPS: 在对mergeSort说两句，在mergeSort中，有条件判断 if(low < high)，但low>= high时，就返回了，即一个元素时就什么也不做，然后执行merge()函数，就自定向上完成了排序。。

 1 class Solution {
 2
 3     void merge(int* array, int low, int mid, int high)
 4     {
 5         int *newArray = (int*)malloc(sizeof(int)* (high-low +1));
 6
 7         int i = low, j = mid+1, k = 0;
 8
 9         while(i <= mid && j <= high)
10         {
11             if(array[i] < array[j])
12             {
13                 newArray[k++] = array[i++];
14             }
15             else
16                 newArray[k++] = array[j++];
17         }
18
19         while(i <= mid)
20             newArray[k++] = array[i++];
21
22         while(j<=high)
23             newArray[k++] = array[j++];
24
25         for(i = low; i<= high; i++)
26             array[i] = newArray[i];
27
28         free(newArray);
29     }
30
31     void mergeSort(int* array, int low, int high)
32     {
33         if(low < high)
34         {
35             int mid = (low + high)/2;
36
37             mergeSort(array, low, mid);
38             mergeSort(array, mid+1, high);
39             merge(array, low, mid, high);
40         }
41     }
42
43 public:
44     vector<vector<int> > threeSum(vector<int> &array) {
45
46         int n = array.size();
47
48         sort(array.begin(), array.end());
49
50          vector<vector<int>> result;
51
52         //mergeSort(array, 0, n-1);
53
54         int a, b , c;
55
56         for(a = 0; a < n-2; )
57         {
58             b = a +1;
59             c = n-1;
60             while(b<c)
61             {
62                 if(array[a] + array[b]+ array[c] == 0)
63                 {
64                     vector<int> temp;
65                     temp.push_back(array[a]);
66                     temp.push_back(array[b]);
67                     temp.push_back(array[c]);
68                     result.push_back(temp);
69                     // this is very important b++ and c--
70
71                     //skip the dupliate ones
72                     do {b++;} while(array[b] == array[b-1] && (b < n-1));
73                     do {c--;} while(array[c] == array[c+1] && (c > a));
74                 }
75                 else if(array[a] + array[b]+ array[c] < 0)
76                     b++;
77                 else
78                     c--;
79             }
80             a++;
81             // skip the duplicate ones
82             while (array[a] == array[a-1] && a < n-2) a++;
83         }
84
85
86
87         return result;
88     }
89 };

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