475. Heaters*

本文介绍了一种算法,用于计算使所有房屋都能被暖气覆盖所需的最小供暖半径。通过整理暖气站和房屋的位置,并利用二分查找确定每个房屋到最近暖气站的距离,最终找出最大距离作为供暖半径。

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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.

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:

  1. Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
  2. Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
  3. As long as a house is in the heaters' warm radius range, it can be warmed.
  4. 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 hous

技巧:

[1] 搜索值不是数组元素,且在数组范围内,从1开始计数,得“ - 插入点索引值”;

[2] 搜索值是数组元素,从0开始计数,得搜索值的索引值;

[3] 搜索值不是数组元素,且大于数组内元素,索引值为 – (length + 1);

[4] 搜索值不是数组元素,且小于数组内元素,索引值为 – 1。


     public int findRadius(int[] houses, int[] heaters) {
        Arrays.sort(heaters);
        int result = Integer.MIN_VALUE;
        for (int house : houses) {
        	int index = Arrays.binarySearch(heaters, house);
        	if (index < 0) {
        		index = -(index + 1);
        	}
        	int dist1 = index - 1 >= 0 ? house - heaters[index - 1] : Integer.MAX_VALUE;
        	int dist2 = index < heaters.length ? heaters[index] - house : Integer.MAX_VALUE;
        	
        	result = Math.max(result, Math.min(dist1, dist2));
        }
        
        return result;
    }

总结:利用了binarysearch找到某house相邻两个heaters

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