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122 lines
3.8 KiB
Markdown
Executable File
122 lines
3.8 KiB
Markdown
Executable File
# [475. Heaters](https://leetcode.com/problems/heaters/)
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## 题目
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Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.
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Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.
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So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.
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**Note**:
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1. Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
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2. Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
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3. As long as a house is in the heaters' warm radius range, it can be warmed.
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4. All the heaters follow your radius standard and the warm radius will the same.
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**Example 1**:
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Input: [1,2,3],[2]
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Output: 1
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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.
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**Example 2**:
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Input: [1,2,3,4],[1,4]
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Output: 1
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Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
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## 题目大意
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冬季已经来临。 你的任务是设计一个有固定加热半径的供暖器向所有房屋供暖。现在，给出位于一条水平线上的房屋和供暖器的位置，找到可以覆盖所有房屋的最小加热半径。所以，你的输入将会是房屋和供暖器的位置。你将输出供暖器的最小加热半径。
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说明:
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- 给出的房屋和供暖器的数目是非负数且不会超过 25000。
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- 给出的房屋和供暖器的位置均是非负数且不会超过10^9。
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- 只要房屋位于供暖器的半径内(包括在边缘上)，它就可以得到供暖。
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- 所有供暖器都遵循你的半径标准，加热的半径也一样。
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## 解题思路
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- 给出一个房子坐标的数组，和一些供暖器坐标的数组，分别表示房子的坐标和供暖器的坐标。要求找到供暖器最小的半径能使得所有的房子都能享受到暖气。
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- 这一题可以用暴力的解法，暴力解法依次遍历每个房子的坐标，再遍历每个供暖器，找到距离房子最近的供暖器坐标。在所有这些距离的长度里面找到最大值，这个距离的最大值就是供暖器半径的最小值。时间复杂度 O(n^2)。
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- 这一题最优解是二分搜索。在查找距离房子最近的供暖器的时候，先将供暖器排序，然后用二分搜索的方法查找。其他的做法和暴力解法一致。时间复杂度 O(n log n)。
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## 代码
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```go
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package leetcode
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import (
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"math"
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"sort"
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)
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func findRadius(houses []int, heaters []int) int {
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minRad := 0
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sort.Ints(heaters)
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for _, house := range houses {
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// 遍历房子的坐标，维护 heaters 的最小半径
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heater := findClosestHeater(house, heaters)
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rad := heater - house
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if rad < 0 {
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rad = -rad
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}
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if rad > minRad {
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minRad = rad
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}
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}
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return minRad
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}
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// 二分搜索
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func findClosestHeater(pos int, heaters []int) int {
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low, high := 0, len(heaters)-1
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if pos < heaters[low] {
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return heaters[low]
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}
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if pos > heaters[high] {
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return heaters[high]
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}
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for low <= high {
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mid := low + (high-low)>>1
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if pos == heaters[mid] {
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return heaters[mid]
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} else if pos < heaters[mid] {
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high = mid - 1
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} else {
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low = mid + 1
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}
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}
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// 判断距离两边的 heaters 哪个更近
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if pos-heaters[high] < heaters[low]-pos {
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return heaters[high]
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}
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return heaters[low]
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}
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// 解法二 暴力搜索
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func findRadius1(houses []int, heaters []int) int {
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res := 0
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for i := 0; i < len(houses); i++ {
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dis := math.MaxInt64
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for j := 0; j < len(heaters); j++ {
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dis = min(dis, abs(houses[i]-heaters[j]))
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}
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res = max(res, dis)
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}
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return res
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}
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``` |