[LeetCode] 849. Maximize Distance to Closest Person

本文介绍了一种解决“最大化距离就坐”问题的方法，通过BFS遍历算法找到空座位中离最近的人最远的位置。文章给出了具体实现代码，并解释了如何通过队列维护当前已知最近人员的距离。

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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.

思路就是类似于[LeetCode] 286. Walls and Gates_Medium tag: BFS

Code: T: O(n) S; O(n)

class Solution:
    def maxDistToClosest(self, seats):
        """
        :type seats: List[int]
        :rtype: int
        """
        queue, ans, visited, lr = collections.deque(), 0, set(), len(seats)
        for i in range(lr):
            if seats[i]:
                queue.append((i,0))
                visited.add(i)
        
        while queue:
            node, dis = queue.popleft()
            for c in [-1,1]:
                nr = node + c
                if 0 <= nr < lr and nr not in visited and not seats[nr]:
                    queue.append((nr, dis+1))
                    visited.add(nr)
                    ans = max(ans, dis+1)
        return ans