3 Sum（3 個(gè)數(shù)的和）

翻譯

給定一個(gè)有 n 個(gè)整數(shù)的數(shù)組 S，是否存在三個(gè)元素 a,b,c 使得 a+b+c=0? 找出該數(shù)組中所有不重復(fù)的 3 個(gè)數(shù)，它們的和為 0。

備注：
這三個(gè)元素必須是從小到大進(jìn)行排序。
結(jié)果中不能有重復(fù)的 3 個(gè)數(shù)。

例如，給定數(shù)組 S={-1 0 1 2 -1 4}，一個(gè)結(jié)果集為：
(-1, 0, 1)
(-1, -1, 2)

原文

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)

經(jīng)典方法，可惜我并沒有想到這樣寫……

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> result;

        int len = nums.size();      
        for (int current = 0; current < len - 2&&nums[current]<=0;current++)
        {
            int front = current + 1, back = len - 1;
            while (front < back)
            {
                if (nums[current] + nums[front] + nums[back] < 0)
                    front++;
                else if (nums[current] + nums[front] + nums[back] > 0)
                    back--;
                else
                {
                    vector<int> v(3);
                        v.push_back(nums[current]);
                        v.push_back(nums[front]);
                        v.push_back(nums[back]);
                        result.push_back(v);
                        v.clear();
                    do {
                        front++;
                    } while (front < back&&nums[front - 1] == nums[front]);
                    do {
                        back--;
                    } while (front < back&&nums[back + 1] == nums[back]);
                }
            }                    
            while (current < len - 2 && nums[current + 1] == nums[current])
                current++;
        }                                  
        return result;
    }
};

繼續(xù)努力……

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