[leetcode] 475. Heaters @ python

原题

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

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.

So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.

Note:

Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
As long as a house is in the heaters’ warm radius range, it can be warmed.
All the heaters follow your radius standard and the warm radius will the same.

Example 1:

Input: [1,2,3],[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: [1,2,3,4],[1,4]
Output: 1
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.

解法

二分搜索法.

对于每一个房屋, 它在heaters列表里有三种情况: 它在列表的左边, 它在列表的右边, 它在列表之间, 对于它在列表之外的情况, 求它到最左边或者最右边的加热器的距离即可, 对于它在列表之间的情况, 求它到左右两边的加热器的距离的较小值. 因此使用bisect.besect_left()求它应该处在列表的位置, 然后构建dp列表, 储存所有house到加热器的最小距离, 然后求dp的最大值即可.

代码

class Solution(object):
    def findRadius(self, houses, heaters):
        """
        :type houses: List[int]
        :type heaters: List[int]
        :rtype: int
        """
        houses.sort()
        heaters.sort()
        dp = []
        for house in houses:
            idx = bisect.bisect_left(heaters, house)
            if idx == 0:
                # house is in the left of heaters
                dp.append(heaters[0] - house)
            elif idx == len(heaters):
                # house is in the right of heaters
                dp.append(house - heaters[-1])
            else:
                # get the min distance between two heaters
                dp.append(min(house - heaters[idx-1], heaters[idx]-house))
                
        return max(dp)