LeetCode(15)题解--3Sum

https://leetcode.com/problems/3sum/

题目：

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)

方法一：

两层循环+二分查找，复杂度O(n^2 logn). 太慢了

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        set<vector<int>> res;
        vector<vector<int>> output;
        vector<int> sol;
        sort(nums.begin(),nums.end());    //sort first
        int i,j,k,t,x,n=nums.size();
        for(i=0;i<n;i++){
            for(j=i+1;j<n-1;j++){
                t=nums[i]+nums[j];
                x=findSol(-t,nums,j+1,n-1);
                if(x!=-1){
                    sol.clear();
                    sol.push_back(nums[i]);
                    sol.push_back(nums[j]);
                    sol.push_back(nums[x]);
                    res.insert(sol);
                }
            }
        }
        set<vector<int>> :: iterator iter;
        for(iter=res.begin();iter!=res.end();iter++){
            output.push_back(*iter);
        }
        return output;
    }
    int findSol(int target,vector<int> nums,int begin,int end){
        if(nums[begin]==target)
            return begin;
        if(nums[end]==target)
            return end;            
        if(nums[begin]>target||nums[end]<target)
            return -1;
        int mid;
        while(begin<=end){
            mid=(begin+end)/2;
            if(nums[mid]==target)
                return mid;
            else if(nums[mid]>target){
                end=mid-1;
            }
            else
                begin=mid+1;
        }
        return -1;
    }
};

方法二：  一层循环加two sum思想（https://leetcode.com/problems/two-sum/）,O(n^2).

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        int i,t,a,b,k,n=nums.size();
        for(i=0;i<n-2;i++){
            t=-nums[i];
            a=i+1;
            b=n-1;
            while(a<b){
                k=nums[a]+nums[b];
                if(k<t){
                    a++;
                }
                else if(k>t){
                    b--;
                }
                else{
                    vector<int> sol(3,0);
                    sol[0]=nums[i];
                    sol[1]=nums[a];
                    sol[2]=nums[b];
                    res.push_back(sol);
                    while (a < b && nums[a] == sol[1]) 
                        a++;
                    while (a < b && nums[b] == sol[2]) 
                        b--;
                 }
            }
            while (i + 1 < nums.size() && nums[i + 1] == nums[i]) 
            i++;
        }
        
        //deduplicate, but it is so slow!
        //sort(res.begin(), res.end());           
        //res.erase(unique(res.begin(), res.end()), res.end());
        return res;
    }
};