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=;i<n;i++){
             for(j=i+;j<n-;j++){
                 t=nums[i]+nums[j];
                 x=findSol(-t,nums,j+,n-);
                 if(x!=-){
                     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 -;
         int mid;
         while(begin<=end){
             mid=(begin+end)/;
             if(nums[mid]==target)
                 return mid;
             else if(nums[mid]>target){
                 end=mid-;
             }
             else
                 begin=mid+;
         }
         return -;
     }
 };

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=;i<n-;i++){
             t=-nums[i];
             a=i+;
             b=n-;
             while(a<b){
                 k=nums[a]+nums[b];
                 if(k<t){
                     a++;
                 }
                 else if(k>t){
                     b--;
                 }
                 else{
                     vector<int> sol(,);
                     sol[]=nums[i];
                     sol[]=nums[a];
                     sol[]=nums[b];
                     res.push_back(sol);
                     while (a < b && nums[a] == sol[])
                         a++;
                     while (a < b && nums[b] == sol[])
                         b--;
                  }
             }
             while (i +  < nums.size() && nums[i + ] == 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;
     }
 };