32-3Sum

相似题目 4sum http://blog.csdn.net/justdoithai/article/details/51195124

http://blog.csdn.net/justdoithai/article/details/51122038 2sum

1. 3Sum My Submissions QuestionEditorial Solution
   Total Accepted: 115154 Total Submissions: 612489 Difficulty: Medium
   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：

311 / 311 test cases passed.

Status: Accepted

Runtime: 76 ms

beats 10.13%

突然发现，一个程序的思路很重要，反映在runtime上，以后贴上runtime来
衡量程序优劣，虽然只是一个小程序，但真正在大数据量的情况相差是很大的
之前用map存好任意两个数的和，慢很多，在以下的一个case中发现
改进后的只要1ms，而用map的需要50ms左右，以下是代码

思路：先固定一个数，其余两个数从两端夹逼，再这个过程自动去重
同时过滤排序中已经判定过的情况
时间复杂度：O(n2)
空间复杂度：O(1)

改进：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int> > res;
        if(nums.size()<3)return res;
        sort(nums.begin(),nums.end());
        int n = nums.size();
        for(int i=0;i<nums.size()-1;++i){
            if(i!=0&&nums[i]==nums[i-1])continue;
            int beg = i+1,end = n-1;
            while(beg<end){
                vector<int> tmp;
                int sum2 = nums[beg]+nums[end];
                if(sum2==-nums[i]){
                    tmp.push_back(nums[i]);
                    tmp.push_back(nums[beg]);
                    tmp.push_back(nums[end]);
                    res.push_back(tmp);
                    while(++beg<end&&nums[beg-1]==nums[beg]);
                    while(--end>beg&&nums[end]==nums[end+1]);
                }
                else{
                    if(sum2<-nums[i])++beg;
                    else --end;
                }
            }
        }
        return res;
    }
};

用map的：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int> > res;
        map<vector<int>,int> mres;
        if(nums.size()<3)return res;
        map<int,vector<pair<int,int> > > Two_map;
        for(int i=0;i<nums.size()-1;++i)
        {
            for(int j=i+1;j<nums.size();++j)
                Two_map[nums[i]+nums[j]].push_back(pair<int,int>(i,j));
        }
        for(int i=0;i<nums.size();++i)
        {
            if(Two_map.count(-nums[i]))
            {
                const auto&vec =Two_map[-nums[i]];
                for(size_t k=0;k<vec.size();++k){
                    if(vec[k].first!=i&&vec[k].second!=i)
                    {
                        vector<int> tmp;
                        tmp.push_back(nums[i]);tmp.push_back(nums[vec[k].first]);tmp.push_back(nums[vec[k].second]);
                        sort(tmp.begin(),tmp.end());
                        mres[tmp]=1;
                    }    
                }
            }
        }
        vector<vector<int> > res1;
        auto mit = mres.begin();
        while(mit!=mres.end()){
            res1.push_back(mit->first);
            mit++;
        }
        return res1;
    }
};