本文主要是介绍3Sum Closest问题及解法,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
问题描述:
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.
示例:
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).
问题分析:
要求三个数的和最接近目标值,可分为以下几个步骤:
1.将数组升序排序
2.设置两个指针 front 和 back 分别指向数组的前段和后端
3.先选取好第一个元素,然后遍历第二和第三个元素,将三者的和与目标值target作比较,选取距离较近的
过程详见代码:
class Solution {
public:int threeSumClosest(vector<int>& nums, int target) {int res = 0;int dist = INT_MAX;sort(nums.begin(), nums.end());//排序 for(int i = 0; i < nums.size(); i++){int front = i + 1;int back = nums.size() - 1;while(front < back){int sum = nums[i] + nums[front] + nums[back];if(sum < target) {int temp = target - sum;if(temp < dist){dist = temp;res = sum;}front++;}else if(sum > target){int temp = sum - target;if(temp < dist){dist = temp;res = sum;}back--;}else{return target;}}}return res;}
};
问题难度不大,有疑问的话欢迎交流~~
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