leetcode 475 Heaters

Problem:

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.

解题思路：

首先对每一个house，计算离之最近的heater的距离。最后取这些距离中最大的。
先将houses和heaters数组排序，然后对heaters数组使用两个指针，一个指向当前，一个指向下一个
遍历houses，当house对当前heater的距离大于等于对下一个heater的距离的时候，指针移动，直到距离小于，或者next指向heater尾部。
在遍历的同时，维护一个max值，当对一个house得到与heart最小的距离的时候，与当前max值比较，若大于，则更新

代码如下：

class Solution {
public:int findRadius(vector<int>& houses, vector<int>& heaters) {
        sort(houses.begin(), houses.end());
        sort(heaters.begin(), heaters.end());
        int startindex = 0, maxn = 0;
        for(int i = 0; i < houses.size(); i++) {
            int tempmin = INT_MAX;
            for(int j = startindex; j < heaters.size(); j++) {
                if(abs(heaters[j] - houses[i]) <= tempmin) {
                    tempmin = abs(heaters[j] - houses[i]);
                    startindex = j;
                } else {
                    break;
                }
            }
            maxn = max(maxn, tempmin);
        }
        return maxn;
    }
};