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記錄0的位置，j記錄1的位置,三種情況都要考慮到位
1. [0,0,0,0,1] 0在開頭
2. [1,0,0,0,1] 0在中間
3. [1,0,0,0,0] 0在結果
 public int maxDistToClosest(int[] seats) {
        if (seats == null || seats.length == 0) {
            
 
return 0;
        }
        
        int i = 0;//記錄1後面第一個零的位置
        int result = 0;
        
        for (int j = 0; j < seats.length; j++) {//記錄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是為了判斷0在開頭的情況，否則是0在中間
                i = j + 1;
            }
        }
        
        return Math.max(result, seats.length - i);//0在結果的情況
    }

849 Maximize Distance to Closest Person