18. 4Sum

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]
]

分析

两层循环，然后按照3Sum的做。O(n3)的复杂度。150ms

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class Solution { public:     vector<vector<int>> fourSum(vector<int>& nums, int target) {         int len = nums.size();         vector<vector<int> > result;         if(len < 4) return result;         sort(nums.begin(), nums.end());         for(int i = 0; i < len - 3; ++i){             if( i > 0 && nums[i] == nums[i - 1]){                 continue;             }             for(int j = i + 1; j < len - 2; ++j){                 if(j > i + 1 && nums[j] == nums[j - 1]){                     continue;                 }                 int l = j + 1;                 int r = len - 1;                 while( l < r){                     if( nums[i] + nums[j] + nums[l] + nums[r] == target){                         result.push_back(vector<int>{nums[i], nums[j], nums[l], nums[r]});                         l++;                         while( l < r && nums[l] == nums[l - 1])                             l++;                     }                     else if( nums[i] + nums[j] + nums[l] + nums[r] < target){                         l++;                     }                     else{                         r--;                     }                 }             }         }         return result;     } };

进一步优化：https://discuss.leetcode.com/topic/37878/c-implementation-with-carefully-pruning-accelerates-a-lot-from-100ms-to-16ms

对排序后的数组，进行计算的时候，如果满足 

1. 最靠前的4个数字之和 > target，则退出计算，因为以后也一定不会满足

2. 前面的数组和 末尾的数字之和 < target，那么说明肯定中间数字之和一定是 < target的，继续

1 2	if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break; if(nums[i]+nums[len-3]+nums[len-2]+nums[len-1]<target) continue;

同时优化满足条件的判断顺序以及后续处理，可以得到如下代码：16ms

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class Solution { public:     vector<vector<int>> fourSum(vector<int>& nums, int target) {         int len = nums.size();         vector<vector<int> > result;         if(len < 4) return result;         sort(nums.begin(), nums.end());         for(int i = 0; i < len - 3; ++i){             if( i > 0 && nums[i] == nums[i - 1]){                 continue;             }             /** cut edge to accelerate the speed **/             if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;             if(nums[i]+nums[len-3]+nums[len-2]+nums[len-1]<target) continue;             for(int j = i + 1; j < len - 2; ++j){                 if(j > i + 1 && nums[j] == nums[j - 1]){                     continue;                 }                 /** cut edge to accelerate the speed **/                 if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;                 if(nums[i]+nums[j]+nums[len-2]+nums[len-1]<target) continue;                 int l = j + 1;                 int r = len - 1;                 while( l < r){                     int sum = nums[i] + nums[j] + nums[l] + nums[r];                     if( sum < target){                         l++;                     }                     else if( sum > target){                         r--;                     }                     else{                         result.push_back(vector<int>{nums[i], nums[j], nums[l], nums[r]});                         /** cut edge to accelerate the speed **/                         r--;l++;                         while( l < r && nums[l] == nums[l - 1])l++;                         while( l < r && nums[r] == nums[r + 1])r--;                     }                 }             }         }         return result;     } };

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