Leetcode: 16. 3Sum Closest(Week5, Medium)
2017-10-05 16:26
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Leetcode 16
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
题意:从给定数组中找到三个数,使其和最接近给出的target
思路:暴力解法,三重for循环遍历,并设定一个数组与一个变量来存储当前的最小差异值以及该值对应的三个数
代码如下:
以上内容皆为本人观点,欢迎大家提出批评和知道,我们一起探讨!
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
题意:从给定数组中找到三个数,使其和最接近给出的target
思路:暴力解法,三重for循环遍历,并设定一个数组与一个变量来存储当前的最小差异值以及该值对应的三个数
代码如下:
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { vector<int> temp(3, 0); int diff = 0x7FFFFFFF; for (int i = 0; i < nums.size(); i++) { for (int j = i+1; j < nums.size(); j++) { for (int k = j+1; k < nums.size(); k++) { int temp_diff = abs(nums[i] + nums[j] + nums[k] - target); if (temp_diff == 0) return nums[i] + nums[j] + nums[k]; if (temp_diff < diff) { diff = temp_diff; temp[0] = nums[i]; temp[1] = nums[j]; temp[2] = nums[k]; } } } } return temp[0] + temp[1] + temp[2]; } };
以上内容皆为本人观点,欢迎大家提出批评和知道,我们一起探讨!
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