LeetCode 18 - 4Sum
2015-05-20 12:49
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一、问题描述
Description: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)
给定一个包含 n 个整数的数组,以及一个目标整数 target,找出其中所有和等于 target 的四元组。
注意:
每个四元组中的4个数按大小顺序排列,a ≤ b ≤ c ≤ d 。
返回的四元组不能重复。
二、解题报告
K sum 是一类问题,在《LeetCode 15 - 3Sum》中说到:3Sum可以通过先固定一个数,从而转化为2Sum。同理,4sum 也可以先固定一个数,转化成3sum问题;然后再固定一个数,最后也就是解决一个2sum的问题。
下面直接上代码:O(n3)
class Solution { vector<vector<int>> res; public: vector<vector<int>> fourSum(vector<int>& nums, int target) { if(nums.size() < 4) return res; sort(nums.begin(), nums.end()); // 排序 for(int i=0; i<nums.size()-3; ++i) { if(i>0 && nums[i]==nums[i-1]) // 相等的跳过 continue; threeSum(nums, i+1, nums[i], target); } return res; } void threeSum(vector<int>& nums, int p, int first, int target) { for(int i=p; i<nums.size()-2; ++i) { if(i>p && nums[i]==nums[i-1]) // 相等的跳过 continue; twoSum(nums, i+1, nums.size()-1, first, nums[i], target); } } void twoSum(vector<int>& nums, int begin, int end, int first, int second, int target) { int i = begin; // 头指针 int j = end; // 尾指针 while(i < j) { if(first+second+nums[i]+nums[j]==target) { vector<int> v; v.push_back(first); v.push_back(second); v.push_back(nums[i]); v.push_back(nums[j]); res.push_back(v); while(i<j && nums[i]==nums[i+1]) ++i; // 相等的跳过 while(i<j && nums[j]==nums[j-1]) --j; // 相等的跳过 ++i; --j; } else if(first+second+nums[i]+nums[j]<target) ++i; else --j; } } };
LeetCode答案源代码:https://github.com/SongLee24/LeetCode
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