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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 | 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; }}; |
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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 | 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; }}; |