Array + two points leetcode.18 - 4Sum
题面
Given an array nums
of n integers and an integer target
, are there elements a, b, c, and d in nums
such that a + b+ c + d = target
? Find all unique quadruplets in the array which gives the sum of target
.
给定无序数组,找到和为target的不重复的长度为4的子序列
样例
1. Given array nums = [1, 0, -1, 0, -2, 2], and target = 0. solution set is: [ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ]
2. Given array nums = [0, 0, 0, 0], and target = 0. solution set is: [ [0, 0, 0, 0] ]
note: This example need to think specially!
3. Given array nums = [-3, -3, -3, 2, 2, 2,0, 0, 0, 3, 3, 3], and target = 0. solution set is: [ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ]
Note: How to deal with the duplicate quadruplets ?
思路(这里是4个数的和为target,我们之前不是做过3个数的吗?Refer this one 正好可以用上)
1. 升序排序(sort即可)
2. 不可避免地要遍历数组(i)
3. 借鉴leetcode-15中的三个数的和,我们如法炮制,搜索剩下的三个数(j = i+1, l=j+1, r = size()-1);对参数有疑惑的话,看下面代码。
时间复杂度:O(n3),暴力的话应该会超时,没试!
这题里面尤其要注意避免重复
源码
1 class Solution { 2 public: 3 vector<vector<int>> fourSum(vector<int>& nums, int target) { 4 vector<vector<int>> res; 5 int len = nums.size(); 6 if(len < 4) 7 return res; 8 9 sort(nums.begin(), nums.end()); 10 for(int i=0; i<=len-4; i++) 11 { 12 while(i>0 && i<=len-4 && nums[i]==nums[i-1]) 13 i++;//消除i的重复 14 for(int j=i+1; j<=len-3; j++) 15 { 16 while(j>i+1 && j<=len-3 && nums[j]==nums[j-1]) 17 j++;//消除j的重复 18 int l=j+1, r=len-1; 19 while(l < r) 20 { 21 int sum = nums[i] + nums[j] + nums[l] + nums[r]; 22 if(sum > target) 23 r--; 24 else if(sum < target) 25 l++; 26 else 27 { 28 res.push_back(vector<int> {nums[i], nums[j], nums[l], nums[r]}); 29 while(l<r && nums[l]==nums[l+1]) l++;//消除l的重复 30 while(l<r && nums[r]==nums[r-1]) r--;//消除r的重复 31 l++; r--; 32 } 33 } 34 } 35 } 36 return res; 37 } 38 };
如果对消除重复有疑问的童鞋,请留言, 或者自行把example 3 手推一遍就明白了。