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96 lines
3.5 KiB
Markdown
96 lines
3.5 KiB
Markdown
# [475. Heaters](https://leetcode.com/problems/heaters/description/)
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# 思路
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计算每一个house与其最近heater的距离dist，在计算所有house的dist的最大值即所求。针对如何求每一个house的dist，有三种思路。
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## 思路一
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将heater排序，对与每一个house h，在heaters中二分查找h，若能找到则dist=0，否则dist应该是h与最接近的heater的距离。
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时间复杂度O(n * nlogn), 空间复杂度O(1)
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## 思路二
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思路类似一，只是不自己写二分查找，而是使用stl中的lower_bound函数进行查找。
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lower_bound(begin_it, end_it, target)返回的是处于两个迭代器begin_it和end_it之间(左闭右开)的元素中不小于target的最小元素的迭代器。若不存在返回end_it.
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> 注意：begin_it到end_it的元素需有序
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复杂度同思路一
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## 思路三*
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如果houses和heaters都是有序的话，其实不用每次都二分查找, 而只需要遍历一次就行了。这是因为我们注意到一个事实：
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若离house h最近的heater为ht，那么离下一个house最近的heater的位置大小肯定不小于ht，所以不用回溯ht。
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时间复杂度O(n^2), 空间复杂度O(1)
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# C++
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## 思路一
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```
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class Solution {
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private:
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int find_max_dist(vector<int>&sorted_heaters, int &h){ # 二分查找h
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int low = 0, high = sorted_heaters.size() - 1, mid;
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while(low <= high){
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mid = low + (high - low) / 2;
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if(sorted_heaters[mid] == h) return 0; // 找到
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if(sorted_heaters[mid] < h) low = mid + 1;
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else high = mid - 1;
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}
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// 没找到，计算最小dist
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if(high < 0) return sorted_heaters[low] - h;
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else if(low >= sorted_heaters.size()) return h - sorted_heaters[high];
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else return min(sorted_heaters[low] - h, h - sorted_heaters[high]);
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}
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public:
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int findRadius(vector<int>& houses, vector<int>& heaters) {
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sort(heaters.begin(), heaters.end());
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int res = 0;
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for(int h: houses)
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res = max(res, find_max_dist(heaters, h));
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return res;
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}
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};
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```
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## 思路二
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```
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class Solution {
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private:
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int find_max_dist(vector<int>&sorted_heaters, int &h){
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auto larger = lower_bound(sorted_heaters.begin(), sorted_heaters.end(), h); // 返回的是一个迭代器
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if(larger == sorted_heaters.end()) return h - *(larger - 1);
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else if(larger == sorted_heaters.begin()) return *larger - h;
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else return min(*larger - h, h - *(larger - 1));
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}
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public:
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int findRadius(vector<int>& houses, vector<int>& heaters) {
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sort(heaters.begin(), heaters.end()); # 传入lower_bound的必须是有序的
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int res = 0;
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for (int h: houses) {
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res = max(res, find_max_dist(heaters, h));
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}
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return res;
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}
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};
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```
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## 思路三*
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```
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class Solution {
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public:
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int findRadius(vector<int>& houses, vector<int>& heaters) {
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if (heaters.size() == 0) {
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return 0;
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}
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sort(houses.begin(), houses.end());
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sort(heaters.begin(), heaters.end());
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int res = 0;
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int ht = 0;
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for (int h: houses) {
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// 找到离h最近的heater，这个heater肯定位于当前heater的右边(或就是当前heater)
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while (ht + 1 < heaters.size() && (abs(heaters[ht + 1] - h) <= abs(heaters[ht] - h))) {
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ht++;
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}
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res = max(res, abs(heaters[ht] - h));
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}
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return res;
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}
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};
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```
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