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 <= 20000seatscontains only 0s or 1s, at least one0, and at least one1.
里面陷进比较多，面试的话还是要自己写完备的测试用例才行。
直接贪心求解。
class Solution(object):
def maxDistToClosest(self, seats):
"""
:type seats: List[int]
:rtype: int
"""
# find max contineous 0, postion
i = 0
l = len(seats)
ans = 0
while i < l:
while i<l and seats[i] == 1:
i += 1
# i == l or seats[i] == 0
start = i
while i<l and seats[i] == 0:
i += 1
# i == l or seats[i] == 1
if start == 0 or i == l:
ans = max(ans, i-start)
else:
if (i-start) & 1: # zero cnt is odd
ans = max(ans, (i-start)/2+1)
else:
ans = max(ans, (i-start-1)/2+1)
#i += 1
return ans
如果是基于计数器的思维，则写起来也还行吧：
public int maxDistToClosest(int[] seats) {
int dist = 0;
int temp = 0;
boolean closed = false;
for(int i=0; i<seats.length; ++i){
if(seats[i] == 0)
++temp;
else{
dist = Math.max(dist, closed ? (int)Math.ceil(temp/2.0) : temp);
closed = true;
temp = 0;
}
}
dist = Math.max(dist, temp);
return dist;
}