0539-最小时间差

Raphael Liu Lv10

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”

方法一:排序

将 timePoints 排序后,最小时间差必然出现在 timePoints 的两个相邻时间,或者 timePoints 的两个首尾时间中。因此排序后遍历一遍 timePoints 即可得到最小时间差。

[sol1-Python3]
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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
[sol1-C++]
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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;
}
};
[sol1-Java]
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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');
}
}
[sol1-C#]
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public class Solution {
public int FindMinDifference(IList<string> timePoints) {
timePoints = timePoints.OrderBy(x => x).ToList();
int ans = int.MaxValue;
int t0Minutes = getMinutes(timePoints[0]);
int preMinutes = t0Minutes;
for (int i = 1; i < timePoints.Count; ++i) {
int minutes = getMinutes(timePoints[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[0] - '0') * 10 + (t[1] - '0')) * 60 + (t[3] - '0') * 10 + (t[4] - '0');
}
}
[sol1-Golang]
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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
}
[sol1-C]
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#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;
}
[sol1-JavaScript]
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var findMinDifference = function(timePoints) {
timePoints.sort();
let ans = Number.MAX_VALUE;
let t0Minutes = getMinutes(timePoints[0]);
let preMinutes = t0Minutes;
for (let i = 1; i < timePoints.length; ++i) {
const minutes = getMinutes(timePoints[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;
};

const getMinutes = (t) => {
return ((t[0].charCodeAt() - '0'.charCodeAt()) * 10 + (t[1].charCodeAt() - '0'.charCodeAt())) * 60 + (t[3].charCodeAt() - '0'.charCodeAt()) * 10 + (t[4].charCodeAt() - '0'.charCodeAt());
}

复杂度分析

  • 时间复杂度:O(n\log n),其中 n 是数组 timePoints 的长度。排序需要 O(n\log n) 的时间。

  • 空间复杂度:O(n) 或 O(\log n)。为排序需要的空间,取决于具体语言的实现。

方法二:鸽巢原理

根据题意,一共有 24 \times 60=1440 种不同的时间。由鸽巢原理可知,如果 timePoints 的长度超过 1440,那么必然会有两个相同的时间,此时可以直接返回 0。

[sol2-Python3]
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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
[sol2-C++]
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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) {
int n = timePoints.size();
if (n > 1440) {
return 0;
}
sort(timePoints.begin(), timePoints.end());
int ans = INT_MAX;
int t0Minutes = getMinutes(timePoints[0]);
int preMinutes = t0Minutes;
for (int i = 1; i < n; ++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;
}
};
[sol2-Java]
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class Solution {
public int findMinDifference(List<String> timePoints) {
int n = timePoints.size();
if (n > 1440) {
return 0;
}
Collections.sort(timePoints);
int ans = Integer.MAX_VALUE;
int t0Minutes = getMinutes(timePoints.get(0));
int preMinutes = t0Minutes;
for (int i = 1; i < n; ++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');
}
}
[sol2-C#]
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public class Solution {
public int FindMinDifference(IList<string> timePoints) {
int n = timePoints.Count;
if (n > 1440) {
return 0;
}
timePoints = timePoints.OrderBy(x => x).ToList();
int ans = int.MaxValue;
int t0Minutes = getMinutes(timePoints[0]);
int preMinutes = t0Minutes;
for (int i = 1; i < n; ++i) {
int minutes = getMinutes(timePoints[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[0] - '0') * 10 + (t[1] - '0')) * 60 + (t[3] - '0') * 10 + (t[4] - '0');
}
}
[sol2-Golang]
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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
}
[sol2-C]
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#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) {
if (timePointsSize > 1440) {
return 0;
}
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;
}
[sol2-JavaScript]
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var findMinDifference = function(timePoints) {
const n = timePoints.length;
if (n > 1440) {
return 0;
}
timePoints.sort();
let ans = Number.MAX_VALUE;
let t0Minutes = getMinutes(timePoints[0]);
let preMinutes = t0Minutes;
for (let i = 1; i < n; ++i) {
const minutes = getMinutes(timePoints[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;
};

const getMinutes = (t) => {
return ((t[0].charCodeAt() - '0'.charCodeAt()) * 10 + (t[1].charCodeAt() - '0'.charCodeAt())) * 60 + (t[3].charCodeAt() - '0'.charCodeAt()) * 10 + (t[4].charCodeAt() - '0'.charCodeAt());
}

复杂度分析

  • 时间复杂度:O(\min(n,C)\log\min(n,C)),其中 n 是数组 timePoints 的长度,C=24 \times 60=1440。由于当 n>C 时直接返回 0,排序时的 n 不会超过 C,因此排序需要 O(\min(n,C)\log\min(n,C)) 的时间。

  • 空间复杂度:O(\min(n,C)) 或 O(\log\min(n,C))。为排序需要的空间,取决于具体语言的实现。

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0539-最小时间差