【Leetcode】15、3Sum && 16、3Sum Closest && 18、4Sum
题目一
Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
思路
如果数组是无序的，很难有比O(N3)更好的办法。除非是用哈希表，在得到两个数之和之后，在表中O(1)时间内寻找第三个数，总体复杂度也是O(n2)，但需要O(n)的空间。
如果先将数组排序O(nlogn)，然后固定第一个数，在后面的部分用TwoSum和数组递增的思想，来搜索。总体时间复杂度为O(nlogn+n2)
需要注意的是最后结果中不能有重复的。
class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> res; int len = nums.size(); if (len <= 2) return res; sort(nums.begin(), nums.end()); int i = 0; while (i < len) { int start = i + 1; int end = len - 1; while (start < end) { if (nums[i] + nums[start] + nums[end] == 0){ res.push_back({nums[i], nums[start], nums[end]}); start++; end--; while (start < end && nums[start] == nums[start-1]) start++; while (start < end && nums[end] == nums[end+1]) end--; } else if (nums[i] + nums[start] + nums[end] > 0) end--; else start++; } i++; // 防止重复，不能删去 while (i < len && nums[i] == nums[i-1]) i++; } return res; } };
class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> res; int len = nums.size(); if (len <= 2) return res; sort(nums.begin(), nums.end()); int i = 0; for (int i = 0; i < len - 2; i++) { if(i>0 && nums[i] == nums[i-1]) // 防止重复 continue; if(nums[i] + nums[i+1] + nums[i+2] > 0) // 减小搜索空间 break; if(nums[i] + nums[len-2] + nums[len-1] + nums[len-1] < 0) // 存在和为的3Sum continue; int start = i + 1; int end = len - 1; while (start < end) { if (nums[i] + nums[start] + nums[end] == 0){ res.push_back({nums[i], nums[start], nums[end]}); start++; end--; while (start < end && nums[start] == nums[start-1]) start++; while (start < end && nums[end] == nums[end+1]) end--; } else if (nums[i] + nums[start] + nums[end] > 0) end--; else start++; } } return res; } };
题目二
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
Example:
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
思路
整体框架和题目一差不多，只需要多一个变量来记录当前的 sum-target 是否比之前的更小即可。
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { int len = nums.size(); int res = nums[0] + nums[1] + nums[len - 1]; sort(nums.begin(), nums.end()); for (int i = 0; i < len - 2; i++) { int start = i + 1, end = len - 1; while (start < end) { int sum = nums[i] + nums[start] + nums[end]; if (sum > target) end--; else start++; if (abs(sum - target) < abs(res - target)) res = sum; } } return res; } };
题目3
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.
Note:
The solution set must not contain duplicate quadruplets.
Example:
Given array nums = [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] ]
思路
和3Sum完全相同，还是转化为2Sum的问题
class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { vector<vector<int>> res; int len = nums.size(); if (len < 4) return res; sort(nums.begin(), nums.end()); for(int i = 0; i < len-3; i++) { // 每次会减少一些搜索空间 if(i > 0 && nums[i] == nums[i-1]) continue; 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++) { // 减小搜索空间 if(j > i+1 && nums[j] == nums[j-1]) continue; 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 left=j + 1,right = len - 1; while(left < right){ int sum = nums[left] + nums[right] + nums[i] + nums[j]; if(sum < target) left++; else if(sum > target) right--; else{ res.push_back({nums[i],nums[j],nums[left],nums[right]}); left++; right--; while(nums[left] == nums[left-1] && left < right) left++; while(nums[right] == nums[right+1] && left < right) right--; } } } } return res; } };
