LeetCode475. Heaters

文章目录

一、题目

Winter is coming! During the contest, your first job is to design a standard heater with a fixed warm radius to warm all the houses.

Every house can be warmed, as long as the house is within the heater's warm radius range.

Given the positions of houses and heaters on a horizontal line, return the minimum radius standard of heaters so that those heaters could cover all houses.

Notice that all the heaters follow your radius standard, and the warm radius will the same.

Example 1:

Input: houses = [1,2,3], heaters = [2]

Output: 1

Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.

Example 2:

Input: houses = [1,2,3,4], heaters = [1,4]

Output: 1

Explanation: The two heaters were placed at positions 1 and 4. We need to use a radius 1 standard, then all the houses can be warmed.

Example 3:

Input: houses = [1,5], heaters = [2]

Output: 3

Constraints:

1 <= houses.length, heaters.length <= 3 * 104

1 <= houses[i], heaters[i] <= 109

二、题解

双指针,i和j分别从houses和heaters的索引从左向右移动。

cpp 复制代码
class Solution {
public:
    int findRadius(vector<int>& houses, vector<int>& heaters) {
        sort(houses.begin(),houses.end());
        sort(heaters.begin(),heaters.end());
        int res = 0;
        for(int i = 0,j = 0;i < houses.size();i++){
            while(!best(houses,heaters,i,j)) j++;
            res = max(res,abs(houses[i]-heaters[j]));
        }
        return res;
    }
    bool best(vector<int>& houses,vector<int>& heaters,int i,int j){
        if(j == heaters.size() - 1) return true;
        else if(abs(houses[i]-heaters[j]) < abs(houses[i]-heaters[j + 1])) return true;
        else return false;
    }
};
相关推荐
木子.李34735 分钟前
排序算法总结(C++)
c++·算法·排序算法
闪电麦坤952 小时前
数据结构:递归的种类(Types of Recursion)
数据结构·算法
freyazzr2 小时前
C++八股 | Day2 | atom/函数指针/指针函数/struct、Class/静态局部变量、局部变量、全局变量/强制类型转换
c++
小熊猫写算法er2 小时前
终极数据结构详解:从理论到实践
数据结构
Gyoku Mint2 小时前
机器学习×第二卷:概念下篇——她不再只是模仿,而是开始决定怎么靠近你
人工智能·python·算法·机器学习·pandas·ai编程·matplotlib
纪元A梦2 小时前
分布式拜占庭容错算法——PBFT算法深度解析
java·分布式·算法
fpcc3 小时前
跟我学c++中级篇——理解类型推导和C++不同版本的支持
开发语言·c++
px不是xp3 小时前
山东大学算法设计与分析复习笔记
笔记·算法·贪心算法·动态规划·图搜索算法
-qOVOp-3 小时前
408第一季 - 数据结构 - 栈与队列的应用
数据结构