一、问题描述
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)1 <= seats.length <= 20000
2)seats contains only 0s or 1s, at least one 0, and at least one 1.
题目要求：
在一排座位中，1表示坐在那个座位上的人，0表示座位是空的。至少有一个空位，至少有一个人坐着。
Alex想坐在座位上，编写一个函数，返回离Alex最近的人的最大距离。
二、思路及代码
思路：
先遍历座位数组seats，分别将0元素和1元素的下标索引保存到数组zero和数组one中；接着计算zero中每个元素与zero中所有元素距离的最小值，并将其保存到数组distance中，最后利用max_element()函数求得的数组distance的最大值即为离Alex最近的人的最大距离。
代码：
int maxDistToClosest(vector<int>& seats)
{int len = seats.size();if (len == 0)return 0;vector<int>one;vector<int>zero;vector<int>distance;for (int i = 0; i < len; i++){if (seats[i] == 0)zero.push_back(i);elseone.push_back(i);}for (int j = 0; j < zero.size(); j++){int min = 1000;for (int z = 0; z < one.size(); z++){ int dis = abs(zero[j] - one[z]);if (min > dis)min = dis; }distance.push_back(min);}vector<int>::iterator max = max_element(distance.begin(), distance.end());return *max;
}