[LeetCode] 849. Maximize Distance to Closest Person 最大化最近人的距离

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.

有一排座位，1代表座位有人，0代表座位空着。最少有1个人和1个空座位。Alex 想坐到一个座位上，使得离他最近的人的距离最大化，返回这个最大距离。

解法：双指针，左指针开始是0， 右指针是循环的index，右指针遇到1就计算与左指针的距离，计算完以后左指针变成现在的index。如果alex坐到两个1中间，则离他最近的人的距离是那两个人的index差除以2。如果第一个和最后一个座位是空位0，则alex可以坐到这个空位上，使得此时的距离最大。然后对所有的距离取最大的返回。

Java:

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

Java:

class Solution {
    public int maxDistToClosest(int[] seats) {
        int left = -1, maxDis = 0;
        int len = seats.length;

        for (int i = 0; i < len; i++) {
            if (seats[i] == 0) continue;

            if (left == -1) {
                maxDis = Math.max(maxDis, i);
            } else {
                maxDis = Math.max(maxDis, (i - left) / 2);
            }
            left = i;
        }

        if (seats[len - 1] == 0) {
            maxDis = Math.max(maxDis, len - 1 - left);
        }

        return maxDis;
    }
}　　

Python:

# Time:  O(n)
# Space: O(1)
class Solution(object):
    def maxDistToClosest(self, seats):
        """
        :type seats: List[int]
        :rtype: int
        """
        prev, result = -1, 1
        for i in xrange(len(seats)):
            if seats[i]:
                if prev < 0:
                    result = i
                else:
                    result = max(result, (i-prev)//2)
                prev = i
        return max(result, len(seats)-1-prev)　　

Python: wo

class Solution(object):
    def maxDistToClosest(self, seats):
        """
        :type seats: List[int]
        :rtype: int
        """
        left = 0
        max_d = 0
        for i in range(len(seats)):
            if seats[i] == 1:
                if seats[0] == 0 and left == 0:
                    max_d = max(max_d, i - left)
                else:
                    max_d = max(max_d, (i - left) / 2)
                left = i

        if seats[-1] == 0:
            max_d = max(max_d, len(seats) - 1 - left)       

        return max_d

C++:　　

int maxDistToClosest(vector<int> seats) {
        int i, j, res = 0, n = seats.size();
        for (i = j = 0; j < n; ++j)
            if (seats[j] == 1) {
                if (i == 0) res = max(res, j - i);
                else res = max(res, (j - i + 1) / 2);
                i = j + 1;
            }
        res = max(res, n - i);
        return res;
    }

　　　　

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