# 849. Maximize Distance to Closest Person 到最近的人的最大距离

@TOC

## # 题目描述

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.

## # 解题方法

1. 设了一个比字符串长度更远的一个字符C，保证后面求最小值更新距离时一定会被更新。
2. 无论如何都用到了abs求绝对值，哪怕可能是不需要的，目的是不用费脑子思考谁大谁小了。

``````class Solution(object):
def maxDistToClosest(self, seats):
"""
:type seats: List[int]
:rtype: int
"""
index = -200000
_len = len(seats)
ans = [0] * _len
for i in range(_len):
if seats[i] == 1:
index = i
else:
ans[i] = abs(i - index)
index = -200000
for i in range(_len - 1, -1, -1):
if seats[i] == 1:
index = i
else:
ans[i] = min(abs(i - index), ans[i])
return max(ans)
``````

## # 日期

2018 年 6 月 10 日 —— 等了两天的腾讯比赛复赛B的数据集，结果人家在复赛刚开始就给了。。 2018 年 11 月 22 日 —— 感恩节快乐～