【LeetCode刷题日记】539. 最小时间差

题目

给定一个 24 小时制（小时:分钟 "HH:MM"）的时间列表，找出列表中任意两个时间的最小时间差并以分钟数表示。

示例 1：
输入：timePoints = ["23:59","00:00"]
输出：1
示例 2：

输入：timePoints = ["00:00","23:59","00:00"]
输出：0

提示：
2 <= timePoints.length <= 2 * 104
timePoints[i] 格式为 "HH:MM"

题解

C++

class Solution {
    int getMinutes(string &t) {
        return (int(t[0] - '0') * 10 + int(t[1] - '0')) * 60 + int(t[3] - '0') * 10 + int(t[4] - '0');
    }

public:
    int findMinDifference(vector<string> &timePoints) {
        sort(timePoints.begin(), timePoints.end());
        int ans = INT_MAX;
        int t0Minutes = getMinutes(timePoints[0]);
        int preMinutes = t0Minutes;
        for (int i = 1; i < timePoints.size(); ++i) {
            int minutes = getMinutes(timePoints[i]);
            ans = min(ans, minutes - preMinutes); // 相邻时间的时间差
            preMinutes = minutes;
        }
        ans = min(ans, t0Minutes + 1440 - preMinutes); // 首尾时间的时间差
        return ans;
    }
};

注意怎么根据 24 小时制（小时:分钟 “HH:MM”）的时间列表这个字符串来获得整型时间。

C

#define MIN(a, b) ((a) < (b) ? (a) : (b))

int getMinutes(const char * t) {
    return ((t[0] - '0') * 10 + (t[1] - '0')) * 60 + (t[3] - '0') * 10 + (t[4] - '0');
}

int cmp(const void * pa, const void * pb) {
    return strcmp(*(char **)pa, *(char **)pb);
}

int findMinDifference(char ** timePoints, int timePointsSize) {
    qsort(timePoints, timePointsSize, sizeof(char *), cmp);
    int ans = 1440;
    int t0Minutes = getMinutes(timePoints[0]);
    int preMinutes = t0Minutes;
    for (int i = 1; i < timePointsSize; ++i) {
        int minutes = getMinutes(timePoints[i]);
        ans = MIN(ans, minutes - preMinutes); // 相邻时间的时间差
        preMinutes = minutes;
    }
    ans = MIN(ans, t0Minutes + 1440 - preMinutes); // 首尾时间的时间差
    return ans;
}

Java

class Solution {
    public int findMinDifference(List<String> timePoints) {
        Collections.sort(timePoints);
        int ans = Integer.MAX_VALUE;
        int t0Minutes = getMinutes(timePoints.get(0));
        int preMinutes = t0Minutes;
        for (int i = 1; i < timePoints.size(); ++i) {
            int minutes = getMinutes(timePoints.get(i));
            ans = Math.min(ans, minutes - preMinutes); // 相邻时间的时间差
            preMinutes = minutes;
        }
        ans = Math.min(ans, t0Minutes + 1440 - preMinutes); // 首尾时间的时间差
        return ans;
    }

    public int getMinutes(String t) {
        return ((t.charAt(0) - '0') * 10 + (t.charAt(1) - '0')) * 60 + (t.charAt(3) - '0') * 10 + (t.charAt(4) - '0');
    }
}