Problem
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).
Solution: Sorting + Two Pointers
Similar to 花花酱 LeetCode 15. 3Sum
Time complexity: O(n^2)
Space complexity: O(1)
C++
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// Author: Huahua, 8 ms class Solution { public: int threeSumClosest(vector<int> &num, int target) { int n = num.size(); int d = INT_MAX; int ans = target; sort(num.begin(), num.end()); for (int i = 0; i < n - 2; i++) { int s = i + 1, t = n - 1; while (s < t) { int sum = num[i] + num[s] + num[t]; if (sum == target) return target; int diff = abs(sum - target); if (diff < d) { d = diff; ans = sum; } if (sum > target) --t; else ++s; } } return ans; } }; |
Java
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// Author: Huahua class Solution { public int threeSumClosest(int[] nums, int target) { Arrays.sort(nums); int n = nums.length; int ans = 0; int d = Integer.MAX_VALUE; for (int i = 0; i < n - 2; ++i) { int s = i + 1; int t = n - 1; while (s < t) { int sum = nums[i] + nums[s] + nums[t]; if (sum == target) return target; int diff = Math.abs(sum - target); if (diff < d) { d = diff; ans = sum; } if (sum > target) --t; else ++s; } } return ans; } } |
Python3
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# Author: Huahua class Solution: def threeSumClosest(self, nums, target): nums.sort() n = len(nums) d = 2**32 ans = 0 for i in range(n - 2): s, t = i + 1, n - 1 while s < t: sum3 = nums[i] + nums[s] + nums[t] if sum3 == target: return target diff = abs(sum3 - target) if diff < d: ans, d = sum3, diff if sum3 > target: t -= 1 else: s += 1 return ans |
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