539 Minimum Time Difference 最小时间差
Description:
Given a list of 24-hour clock time points in "HH:MM" format, return the minimum minutes difference between any two time-points in the list.
Example:
Example 1:
Input: timePoints = ["23:59","00:00"]
Output: 1
Example 2:
Input: timePoints = ["00:00","23:59","00:00"]
Output: 0
Constraints:
2 <= timePoints <= 2 * 104
timePoints[i] is in the format "HH:MM".
题目描述:
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 :
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 <= 2 * 104
timePoints[i] 格式为 "HH:MM"
思路:
转化为分钟数
注意到两头的值可能最小, 可以初始化赋值
排序之后比较差值即可
时间复杂度 O(n), 空间复杂度 O(n)
代码:
C++:
class Solution
{
public:
int findMinDifference(vector& timePoints)
{
int n = timePoints.size();
if (n >= 1440) return 0;
vector count(n);
for (int i = 0; i < n; i++) count[i] = [](const auto &s) { return s[0] * 600 + s[1] * 60 + s[3] * 10 + s[4]; }(timePoints[i]);
sort(count.begin(), count.end());
int m = 1440 + count.front() - count.back();
for (int i = 0; i < n - 1; i++)
{
m = min(m, count[i + 1] - count[i]);
if (!m) return m;
}
return m;
}
};
Java:
class Solution {
public int findMinDifference(List timePoints) {
int n = timePoints.size();
if (n >= 1440) return 0;
int[] count = new int[n];
for (int i = 0; i < n; i++) count[i] = timePoints.get(i).charAt(0) * 600 + timePoints.get(i).charAt(1) * 60 + timePoints.get(i).charAt(3) * 10 + timePoints.get(i).charAt(4);
Arrays.sort(count);
int m = 1440 + count[0] - count[n - 1];
for (int i = 1; i < n; i++) {
m = Math.min(m, count[i] - count[i - 1]);
if (m == 0) return m;
}
return m;
}
}
Python:
class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
start, end, count = 1441, -1, [0] * 1440
for t in timePoints:
t = t.split(":")
count[(t := int(t[0]) * 60 + int(t[1]))] += 1
if count[t] == 2:
return 0
start, end = min(start, t), max(end, t)
result, pre = 1440 + start - end, start
for i in range(start + 1, end + 1):
if count[i]:
result, pre = min(result, i - pre), i
return result