Maximize Distance to Closest Person (M)

题目

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 * 10^4
• seats[i] is 0 or 1.
• At least one seat is empty.
• At least one seat is occupied.

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题意

给定n个座位，其中有至少1个至多n-1个坐着人。将Alex安排到其中一个空座位中，使得他离最近的人的距离最大。

思路

建两个数组left和right。left[i]表示位置i到左边有人的座位的距离，right[i]表示位置i到右边有人的座位的距离。一次遍历生成left和right数组后，在遍历一次比较出答案。

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代码实现

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;
    }
}