18. 4Sum

题目

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.A solution set is:[  [-1,  0, 0, 1],  [-2, -1, 1, 2],  [-2,  0, 0, 2]]

翻译

给定n个整数的数组S，是否在 数组S中有元素a，b，c，d，使得a + b + c + d = target？在数组中找出独一无二的四元素组，使得他们之和为target。

分析

1.一般需要查找的题目都要想一想是否需要排序，此题是需要的

2.通过确定前两个元素，然后二分查找左右逼近，主要在于边界条件的判断。

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {         vector<vector<int>> total;         vector<int> path;         int n=nums.size();         if(n<4)  return total;         sort(nums.begin(),nums.end());         for(int i=0;i<n-3;i++){             if(i>0&&nums[i]==nums[i-1])   continue;             if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target)  break;             if(nums[i]+nums[n-3]+nums[n-2]+nums[n-1]<target)  continue;             for(int j=i+1;j<n-2;j++){                 if(j>i+1&&nums[j]==nums[j-1])   continue;                 if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target)   break;                 if(nums[i]+nums[j]+nums[n-2]+nums[n-1]<target)   continue;                 int left=j+1,right=n-1;                 while(left<right){                     int sum=nums[i]+nums[j]+nums[left]+nums[right];                     if(sum<target)  left++;                     else if(sum>target)  right--;                     else {                                                  total.push_back(vector<int>{nums[i],nums[j],nums[left],nums[right]});                         do {left++ ; }  while(nums[left]==nums[left-1]&&left<right);                         do {right--;  }while(nums[right]==nums[right+1]&&left<right);                 }             }         }             }    return total;    }};