849 Maximize Distance to Closest Person

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

//Two Pointer: Time: O(n), Space: O(1) 
//i记录0的位置，j记录1的位置,三种情况都要考虑到位
1. [0,0,0,0,1] 0在开头
2. [1,0,0,0,1] 0在中间
3. [1,0,0,0,0] 0在结果
 public int maxDistToClosest(int[] seats) {
        if (seats == null || seats.length == 0) {
            return 0;
        }
        
        int i = 0;//记录1后面第一个零的位置
        int result = 0;
        
        for (int j = 0; j < seats.length; j++) {//记录1的位置
            if (seats[j] == 1) {
                result = (i == 0 ? Math.max(result, j - i) : Math.max(result, (j - i + 1) / 2));// i == 0是为了判断0在开头的情况，否则是0在中间
                i = j + 1;
            }
        }
        
        return Math.max(result, seats.length - i);//0在结果的情况
    }