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;      } };

来自为知笔记(Wiz)

转载于:https://www.cnblogs.com/zhxshseu/p/cac3f860d0535ce3571c5cc9e293e5ce.html