# 849.Maximize Distance to Closest Person

**849. Maximize Distance to Closest Person**

Difficulty: Easy

> In a row of seats, 1 represents a person sitting in that seat, and 0 represents that the seat is empty. There is at least one empty seat, and at least one person sitting. Alex wants to sit in the seat such that the distance between him and the closest person to him is maximized. Return that maximum distance to closest person.

Example 1:

```
Input: [1,0,0,0,1,0,1]
Output: 2
Explanation:
If Alex sits in the second open seat (seats[2]), then the closest person has distance 2.
If Alex sits in any other open seat, the closest person has distance 1.
Thus, the maximum distance to the closest person is 2.
```

Example 2:

```
Input: [1,0,0,0]
Output: 3
Explanation:
If Alex sits in the last seat, the closest person is 3 seats away.
This is the maximum distance possible, so the answer is 3.
Note:

1 <= seats.length <= 20000
seats contains only 0s or 1s, at least one 0, and at least one 1.
```

此题类似于 [Leetcode 855](https://github.com/newdee/Leetcode/blob/master/855.Exam_Room.md) 方法： 令距离最近的人的距离最长。先统计有人的座位，然后比较每两个人中间的位置与最近人的距离，最大值纪委所求。\
需要注意开头和结尾的位置。

```
class Solution {
public:
    int maxDistToClosest(vector<int>& seats) {
        vector<int> people;
        int num=seats.size();
        for(int i=0;i<num;i++)
            if(seats[i]) people.push_back(i);

        int d=max(people[0],num-1-people[people.size()-1]);
        for(int i=0;i<people.size()-1;i++)
            d=max(d,(people[i+1]-people[i])/2);
        return d;
    }
};
```


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