Leetcode 849 Maximize Distance to Closest Person

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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.

分析：

  1. 题目用一个列表代表一系列座位，1代表有人，0代表没人，现在要你找一个座位使得离你最近的人尽量远。
  2. 数据长度达到20000，显然n^2方法不行，不过这道题怎么想都想不到n^2上去吧。。

思路：
用一个列表记录下那些位置有人坐，如[1,0,0,0,1,0,1]则记录[0,4,6]，分别作差判断能得到的最远距离是多少，最后处理一下边界问题即可

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def maxDistToClosest(seats):
	rec = []
	for i in range(len(seats)):
		if seats[i] == 1:
			rec.append(i)
	# 处理边界情况
	left = rec[0]
	right = len(seats)-1 - rec[-1]
	res = max(left, right)
	# 依次作差 除2得最远距离
	for i in range(1,len(rec)):
		dis = (rec[i] - rec[i-1]) // 2
		res = max(res,dis)
	return res

79 / 79 test cases passed.
difficulty: easy
Runtime: 65 ms