Leetcode：475. Heaters （利用二分binary search）

題目：
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

先將取暖器數組排序，在遍歷所有house，對每個house，在取暖器數組中進行binary search，如果命中，則說明取暖器位置和house位置重合，這個house的最小半徑爲0；如果沒有命中，則使用返回的index，將index左邊和右邊的取暖器座標與house座標求差值，找出這個house最小半徑。說白了，也是在查找house的最近左右取暖器。
代碼如下：

public static int findRadius(int[] houses, int[] heaters) {
        Arrays.sort(heaters);  
        int result = 0;  
        for(int house:houses) {  
            int index = Arrays.binarySearch(heaters, house);  
            if(index<0) { //index<0,則說明在headers中沒有該house,返回 (-(插入點) - 1),第一個大於此鍵的元素索引    
                index = ~index;  
                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));  
            }  
            //index>=0,則說明在headers中找到了該house, 則說明取暖器位置和house位置重合，該house的最小半徑爲0,result不變  
        }  
        return result;  
    }