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
   .

有一排座位，1代表座位有人，0代表座位空著。最少有1個人和1個空座位。Alex 想坐到一個座位上，使得離他最近的人的距離最大化，返回這個最大距離。

解法：雙指標，左指標開始是0， 右指標是迴圈的index，右指標遇到1就計算與左指標的距離，計算完以後左指標變成現在的index。如果alex坐到兩個1中間，則離他最近的人的距離是那兩個人的index差除以2。如果第一個和最後一個座位是空位0，則alex可以坐到這個空位上，使得此時的距離最大。然後對所有的距離取最大的返回。

Java:

public int maxDistToClosest(int[] seats) {
        int i, j, res = 0, n = seats.length;
        for (i = j = 0; j < n; ++j)
            if (seats[j] == 1) {
                if (i == 0) res = Math.max(res, j - i);
                else res = Math.max(res, (j - i + 1) / 2);
                i = j + 1;
            }
        res = Math.max(res, n - i);
        return res;
    }

Java:

class Solution {
    public int maxDistToClosest(int[] seats) {
        int left = -1, maxDis = 0;
        int len = seats.length;
        
        for (int i = 0; i < len; i++) {
            if (seats[i] == 0) continue;

            if (left == -1) {
                maxDis = Math.max(maxDis, i);
            } else {
                maxDis = Math.max(maxDis, (i - left) / 2);
            }
            left = i;
        }
        
        if (seats[len - 1] == 0) {
            maxDis = Math.max(maxDis, len - 1 - left);
        }
        
        return maxDis;
    }
}　　

Python:

# Time:  O(n)
# Space: O(1)
class Solution(object):
    def maxDistToClosest(self, seats):
        """
        :type seats: List[int]
        :rtype: int
        """
        prev, result = -1, 1
        for i in xrange(len(seats)):
            if seats[i]:
                if prev < 0:
                    result = i
                else:
                    result = max(result, (i-prev)//2)
                prev = i
        return max(result, len(seats)-1-prev)　　

Python: wo

class Solution(object):
    def maxDistToClosest(self, seats):
        """
        :type seats: List[int]
        :rtype: int
        """
        left = 0
        max_d = 0
        for i in range(len(seats)):
            if seats[i] == 1:
                if seats[0] == 0 and left == 0:
                    max_d = max(max_d, i - left)
                else:
                    max_d = max(max_d, (i - left) / 2)
                left = i

        if seats[-1] == 0:
            max_d = max(max_d, len(seats) - 1 - left)       

        return max_d

C++:　　

int maxDistToClosest(vector<int> seats) {
        int i, j, res = 0, n = seats.size();
        for (i = j = 0; j < n; ++j)
            if (seats[j] == 1) {
                if (i == 0) res = max(res, j - i);
                else res = max(res, (j - i + 1) / 2);
                i = j + 1;
            }
        res = max(res, n - i);
        return res;
    }

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