leetcode 849. Maximize Distance to Closest Person（分析复杂）_alex wants to sit in the seat such that the distan-优快云博客

本文探讨了一种算法问题，即在一排座位中找到一个位置，使得某人与最近的已就座者之间的距离最大化。通过分析座位布局，文章提出了解决方案，包括计算两个已就座者之间的空位数、开头和结尾的空位数，以确定最大距离。

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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.

题意：

有一排座位，1代表有人坐，0代表空，此时Alex要选择一个座位，求他离最近的人的最大距离。

思路：

分三种情况分析

两个1夹好多0。计算0的个数除2
以0开头。计算到第一个1之前0的个数。
结尾为0。计算末尾0的个数。

技巧：

通过Math.max()可以始终保存距离最大值

知识点：

a = b == 0 ? 1 : 2; java 中三目运算，分解成 b==0为真则a=1;为假则a=2。

class Solution {
    public int maxDistToClosest(int[] seats) {
        int len = seats.length;
        int dis = 0;
        int ans = 0;
        for(int i=0;i<len;i++){
            dis = seats[i]==1?0:++dis;
            ans = Math.max(ans,(dis+1)/2);
        }
        for(int i=0;i<len;i++){
            if(seats[i]==1){
                ans = Math.max(ans,i);
                break;
            }
        }
        for(int i = len-1;i>=0;i--){
            if(seats[i] == 1){
                ans = Math.max(ans,len-1-i);
                break;
            }
        }
        return ans;
    }
}