849. Maximize Distance to Closest Person-优快云博客

本文链接：https://blog.youkuaiyun.com/BigFatSheep/article/details/84342891

本文解析了一道算法题目，旨在找到在一系列座位中，如何坐下才能使自己与最近的人之间的距离最大。通过计数连续空座位的方法，文章详细阐述了如何在前缀、中间和后缀位置找到最优解，最终实现O(n)的时间复杂度和O(1)的空间复杂度。

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Description

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.

Problem URL

Solution

给一个数组表示座位，1有人坐，0没人坐，保证至少有一个空座位和至少有一个人，问坐下来离最近的人的最远距离。

Approach1:
Count the numbers of continuous zeros in the prefix, res = max(res, zeros)
Count the numbers of continuous zeros in middle, res = max(res, (zeros + 1) / 2)
Count the numbers of continuous zeros in the suffix, res = max(res, zeros)

Code

class Solution {
    public int maxDistToClosest(int[] seats) {
        int i = 0, j = 0, res = 0;
        int n = seats.length;
        for (; j < n; j++){
            if (seats[j] == 1){
                if (i == 0){
                    res = Math.max(res, j);
                }
                else{
                    res = Math.max(res, (j - i + 1) / 2);
                }
                i = j + 1;
            }
        }
        res = Math.max(res, n - i);
        return res;
    }
}