849. Maximize Distance to Closest Person - easy
문제
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.
제한사항
- 1 <= seats.length <= 20000
- seats contains only 0s or 1s, at least one 0, and at least one 1.
입출력 예
1
2
3
4
5
6
7
8
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.
1
2
3
4
5
6
7
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.
풀이
- Array
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
func MaxVal(a int, b int) int {
if(a >= b) {
return a
} else {
return b
}
}
func MinVal(a int, b int) int {
if(a <= b) {
return a
} else {
return b
}
}
func maxDistToClosest(seats []int) int {
res := 0
for i, v := range(seats) {
if(v == 1) {
continue
}
countLeft := 200000
countRignt := 200000
// 현재 인덱스의 -1 부터 처음 인덱스 까지,
// 1이 나올때 까지 검사
for j := i-1 ; j >= 0 ; j-- {
if(seats[j] == 1){
countLeft = i - j
break;
}
}
// 현재 인덱스의 +1 부터 마지막 인덱스 까지,
// 1이 나올때 까지 검사
for j := i+1 ; j < len(seats) ; j++ {
if(seats[j] == 1){
countRignt = j - i
break;
}
}
res = MaxVal(res, MinVal(countLeft, countRignt))
}
return res
}