4Sum
作者:互联网
题目描述:
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:
- Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
- 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)
原理同Two Sum,直接上代码:
solution:
vector<vector<int> > fourSum(vector<int> &num, int target) { int n = num.size(); vector<vector<int> > res; if(n < 4) return res; sort(num.begin(),num.end()); vector<int> temp(4); int low, high; int sum; for (int i = 0;i < n-3;++i) { for (int j = i+1;j < n-2;++j) { sum = target - num[i] - num[j]; low = j + 1; high = n - 1; while (low < high) { if (num[low] + num[high] == sum) { temp[0] = num[i]; temp[1] = num[j]; temp[2] = num[low]; temp[3] = num[high]; res.push_back(temp); ++low; --high; while (low < high && num[low] == num[low-1]) ++low; while (low < high && num[high] == num[high+1]) --high; } else if (num[low] + num[high] < sum) ++low; else --high; } while (j+1 < n-2 && num[j+1] == num[j]) ++j; } while (i+1 < n-3 && num[i+1] == num[i]) ++i; } return res; }
总结:
Two Sum: http://www.cnblogs.com/gattaca/p/4122041.html
3Sum: http://www.cnblogs.com/gattaca/p/4281581.html
3Sum Closest: http://www.cnblogs.com/gattaca/p/4284708.html
kSum: http://tech-wonderland.net/blog/summary-of-ksum-problems.html
转载于:https://www.cnblogs.com/gattaca/p/4285429.html
标签:4Sum,temp,++,high,int,num,low 来源: https://blog.csdn.net/weixin_34226706/article/details/94309047