cpp 解法, 执行用时: 16 ms, 内存消耗: 16.7 MB, 提交时间: 2023-08-22 09:13:16

class Solution {
public:
    int maxDistToClosest(vector<int>& seats) {
        int res = 0;
        int l = 0;
        while (l < seats.size() && seats[l] == 0) {
            ++l;
        }
        res = max(res, l);
        while (l < seats.size()) {
            int r = l + 1;
            while (r < seats.size() && seats[r] == 0) {
                ++r;
            }
            if (r == seats.size()) {
                res = max(res, r - l - 1);
            } else {
                res = max(res, (r - l) / 2);
            }
            l = r;
        }
        return res;
    }
};

php 解法, 执行用时: 40 ms, 内存消耗: 21.6 MB, 提交时间: 2023-08-22 09:12:14

class Solution {

    /**
     * @param Integer[] $seats
     * @return Integer
     */
    function maxDistToClosest($seats) {
        $res = 0;
        $l = 0;
        $n = count($seats);
        // 左侧连续空位个数
        while ( $l < $n && $seats[$l] == 0 ) {
            $l++;
        }
        $res = max($res, $l);
        while ( $l < $n ) {
            $r = $l + 1;
            // 跳过一个有人的，下一个有人的位置
            while ( $r < $n && $seats[$r] == 0 ) {
                $r++;
            }
            if ( $r == $n ) {
                $res = max($res, $r - $l - 1);
            } else {
                $res = max($res, intval(($r - $l) / 2));
            }
            $l = $r;
        }
        return $res;
    }
}

java 解法, 执行用时: 2 ms, 内存消耗: 43.6 MB, 提交时间: 2023-08-22 09:06:02

class Solution {
    public int maxDistToClosest(int[] seats) {
        int res = 0;
        int l = 0;
        while (l < seats.length && seats[l] == 0) {
            ++l;
        }
        res = Math.max(res, l);
        while (l < seats.length) {
            int r = l + 1;
            while (r < seats.length && seats[r] == 0) {
                ++r;
            }
            if (r == seats.length) {
                res = Math.max(res, r - l - 1);
            } else {
                res = Math.max(res, (r - l) / 2);
            }
            l = r;
        }
        return res;
    }
}

golang 解法, 执行用时: 4 ms, 内存消耗: 5.1 MB, 提交时间: 2023-08-22 09:05:37

func maxDistToClosest(seats []int) int {
    res := 0
    l := 0
    for l < len(seats) && seats[l] == 0 {
        l++
    }
    res = max(res, l)
    for l < len(seats) {
        r := l + 1
        for r < len(seats) && seats[r] == 0 {
            r++
        }
        if r == len(seats) {
            res = max(res, r - l - 1)
        } else {
            res = max(res, (r - l) / 2)
        }
        l = r
    }
    return res
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

python3 解法, 执行用时: 64 ms, 内存消耗: 16.8 MB, 提交时间: 2023-08-22 09:05:06

class Solution:
    def maxDistToClosest(self, seats: List[int]) -> int:
        res, l = 0, 0
        while l < len(seats) and seats[l] == 0:
            l += 1
        res = max(res, l)
        while l < len(seats):
            r = l + 1
            while r < len(seats) and seats[r] == 0:
                r += 1
            if r == len(seats):
                res = max(res, r - l - 1)
            else:
                res = max(res, (r - l) // 2)
            l = r
        return res