Maximize Distance to Closest Person (M)
題目
You are given an array representing a row of seats where seats[i] = 1 represents a person sitting in the ith seat, and seats[i] = 0 represents that the ith seat is empty (0-indexed).
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 the closest person.
Example 1:
Input: seats = [1,0,0,0,1,0,1]
Output: 2
Explanation:
If Alex sits in the second open seat (i.e. 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: seats = [1,0,0,0]
Output: 3
Explanation:
If Alex sits in the last seat (i.e. seats[3]), the closest person is 3 seats away.
This is the maximum distance possible, so the answer is 3.
Example 3:
Input: seats = [0,1]
Output: 1
Constraints:
2 <= seats.length <= 2 * 10^4seats[i]is0or1.- At least one seat is empty.
- At least one seat is occupied.
題意
給定n個座位，其中有至少1個至多n-1個坐著人，將Alex安排到其中一個空座位中，使得他離最近的人的距離最大，
思路
建兩個陣列left和right，left[i]表示位置i到左邊有人的座位的距離，right[i]表示位置i到右邊有人的座位的距離，一次遍歷生成left和right陣列后，在遍歷一次比較出答案，
代碼實作
Java
class Solution {
public int maxDistToClosest(int[] seats) {
int[] left = new int[seats.length];
int[] right = new int[seats.length];
int p = -seats.length, q = 2 * seats.length;
for (int i = 0; i < seats.length; i++) {
int j = seats.length - 1 - i;
if (seats[i] == 1) {
p = i;
left[i] = 0;
} else {
left[i] = i - p;
}
if (seats[j] == 1) {
q = j;
right[j] = 0;
} else {
right[j] = q - j;
}
}
int max = 0;
for (int i = 0; i < seats.length; i++) {
max = Math.max(max, Math.min(left[i], right[i]));
}
return max;
}
}
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