[Leetcode 45] 15 3 Sum

Problem:

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)

Analysis:

It's very much like 2 sum problem. 

First need to sort the given array, then pick every different element n in the sorted array, do the following:

1. set two pointers i and j, i points to the next position of n, j points to the lasted element of array

2. get the sum of a[i], a[j], a[n]: if sum == 0, find a solution; if sum > 0, we decrease j by 1; if sum < 0, we increase i by 1; repeat until i > j;

3. after find a solution, we need to update i and j, due to the repeatness of the element, we must find the next element that is different from the current one.

4. update the element n also requires a different new element.

Code:

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() 
        vector<vector<int> > r;
        sort(num.begin(), num.end());
        
        int n = 0;
        while (n < num.size()) {    
            int i = n+1, j = num.size()-1;
            while (i < j) {
                int sum = num[i] + num[j] + num[n];
                
                if (sum == 0) {
                    vector<int> res;
                    res.push_back(num[n]);
                    res.push_back(num[i]);
                    res.push_back(num[j]);
                    r.push_back(res);
                    
                    do {i++;} while(num[i-1] == num[i] && i<num.size());
                    do {j--;} while(num[j] == num[j+1] && j>n);
                    
                } else if (sum > 0) {
                    j--;
                } else if (sum < 0) {
                    i++;
                }
            }
            
            n++;
            while (num[n] == num[n-1] && n < num.size()) n++;
            
            if (n == num.size())
                break;
        }
        
        return r;
    }
};

Attention: