leetcode 3sum

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数组的3Sum

3Sum

解题
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)

题目大意:给定数组，求出数组中三个数的和为0的所有子数组，要求子数组的索引按递增排列。

解题思路:
(1)先将数组排序
(2)固定好第一个数字(外层循环)
(3)第二步就是求2Sum的过程，但需要注意的是判断重复，此处的重复指的是数字重复

先去重，再判断是否满足条件，这样可以减少一次去重(最后一个数)

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int> > result;
        
        if(nums.size() < 3)
            return result;
        
        //先排序
        sort(nums.begin(), nums.end());
        
        for(int i = 0; i < nums.size() - 2; i++){
            //第一个数的去重
            if(i > 0 && nums[i - 1] == nums[i])
                continue;
            
            //two sum
            int target = -nums[i];
            int start = i + 1;
            int end = nums.size() -1;
            while(start < end){
                //跳过计算过的重复的选项(此处去重时，start>i+1)
                if(start > i +1 && nums[start - 1] == nums[start]){
                    start++;
                    continue;
                }
                
                if(nums[start] + nums[end] < target)
                    start++;
                else if(nums[start] + nums[end] > target)
                    end--;
                else{
                    vector<int> solution;
                    solution.push_back(nums[i]);
                    solution.push_back(nums[start]);
                    solution.push_back(nums[end]);
                    
                    result.push_back(solution);
                    start++;
                }
            }
        }
        return result;
    }
};

另一种解法:
先判断是否满足条件，满足条件，再去重(三次)
http://bangbingsyb.blogspot.com/2014/11/leetcode-3sum.html