leetcode: 3Sum

http://oj.leetcode.com/problems/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)

思路：

1. 先排序，然后对于a[i]，找到后面所有令a[i] + a[j] + a[k] = 0的组合。
2. 注意所有的三元组不能有重复，这就需要在循环里做一些处理把重复的数字跳过。

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > result;
        int size = num.size(), tmp;
        
        sort(num.begin(), num.end());
        
        for (int i = 0; i < size; ) {
            int j = i + 1, k = size - 1;
            
            while (j < k) {
                int sum = num[i] + num[j] + num[k];
                
                if (0 == sum) {
                    vector<int> v;
                    v.push_back(num[i]);
                    v.push_back(num[j]);
                    v.push_back(num[k]);
                    result.push_back(v);
                }
                
                if (sum > 0) {
                    tmp = k - 1;
                    while ((j < tmp) && (num[tmp] == num[k])) {
                        --tmp;
                    }
                    k = tmp;
                }
                else {
                    tmp = j + 1;
                    while ((tmp < k) && (num[j] == num[tmp])) {
                        ++tmp;
                    }
                    j = tmp;
                }
            }
            
            tmp = i + 1;
            
            while ((tmp < size) && (num[tmp] == num[i])) {
                ++tmp;
            }
            i = tmp;
        }
        
        return result;
    }
};