class Solution {
public:
int findMinDifference(vector<string>& timePoints) {
}
};
剑指 Offer II 035. 最小时间差
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 <= 2 * 104timePoints[i] 格式为 "HH:MM"
注意：本题与主站 539 题相同： https://leetcode.cn/problems/minimum-time-difference/
原站题解
python3 解法, 执行用时: 32 ms, 内存消耗: 16.6 MB, 提交时间: 2022-11-20 17:36:54
def getMinutes(t: str) -> int:
return ((ord(t[0]) - ord('0')) * 10 + ord(t[1]) - ord('0')) * 60 + (ord(t[3]) - ord('0')) * 10 + ord(t[4]) - ord('0')
class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
n = len(timePoints)
if n > 1440:
return 0
timePoints.sort()
ans = float('inf')
t0Minutes = getMinutes(timePoints[0])
preMinutes = t0Minutes
for i in range(1, n):
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
golang 解法, 执行用时: 4 ms, 内存消耗: 4 MB, 提交时间: 2022-11-20 17:36:04
func getMinutes(t string) int {
return (int(t[0]-'0')*10+int(t[1]-'0'))*60 + int(t[3]-'0')*10 + int(t[4]-'0')
}
func findMinDifference(timePoints []string) int {
if len(timePoints) > 1440 {
return 0
}
sort.Strings(timePoints)
ans := math.MaxInt32
t0Minutes := getMinutes(timePoints[0])
preMinutes := t0Minutes
for _, t := range timePoints[1:] {
minutes := getMinutes(t)
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
}
func min(a, b int) int {
if a > b {
return b
}
return a
}
golang 解法, 执行用时: 8 ms, 内存消耗: 4 MB, 提交时间: 2022-11-20 17:35:40
func getMinutes(t string) int {
return (int(t[0]-'0')*10+int(t[1]-'0'))*60 + int(t[3]-'0')*10 + int(t[4]-'0')
}
func findMinDifference(timePoints []string) int {
sort.Strings(timePoints)
ans := math.MaxInt32
t0Minutes := getMinutes(timePoints[0])
preMinutes := t0Minutes
for _, t := range timePoints[1:] {
minutes := getMinutes(t)
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
}
func min(a, b int) int {
if a > b {
return b
}
return a
}
python3 解法, 执行用时: 60 ms, 内存消耗: 16.5 MB, 提交时间: 2022-11-20 17:35:27
def getMinutes(t: str) -> int:
return ((ord(t[0]) - ord('0')) * 10 + ord(t[1]) - ord('0')) * 60 + (ord(t[3]) - ord('0')) * 10 + ord(t[4]) - ord('0')
class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
timePoints.sort()
ans = float('inf')
t0Minutes = getMinutes(timePoints[0])
preMinutes = t0Minutes
for i in range(1, len(timePoints)):
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