Given a list of 24-hour clock time points in "Hour:Minutes" format, find the minimumminutesdifference between any two time points in the list.

Example 1:

Input: ["23:59","00:00"]
Output: 1

Note:

1. The number of time points in the given list is at least 2 and won't exceed 20000.
2. The input time is legal and ranges from 00:00 to 23:59.

最小時間差。題意是給一些以字串list表示的時間，請你找出這中間最小的時間差。例子給的不是很好，因為list不止兩個元素。

這道題不涉及演算法，不過最優解比較考驗你對時間的敏感度。首先，既然時間是在00:00 - 23:59這個範圍內被表示出來，所以最壞也就是1440種情況，所以可以建立一個長度為1440的布林陣列記錄哪些時間出現過。之後遍歷這個布林陣列，看看哪兩個時間之間的時間差最小。注意記得去“模”24小時。

時間O(n)

空間O(n)

Java實現

 1 class Solution {
 2     public int findMinDifference(List<String> timePoints) {
 
 3         boolean[] mark = new boolean[24 * 60];
 4         for (String time : timePoints) {
 5             String[] t = time.split(":");
 6             int h = Integer.parseInt(t[0]);
 7             int m = Integer.parseInt(t[1]);
 8             if (mark[h * 60 + m]) {
 9                 return 0;
10             }
 
11             mark[h * 60 + m] = true;
12         }
13 
14         int prev = 0;
15         int min = Integer.MAX_VALUE;
16         int first = Integer.MAX_VALUE;
17         int last = Integer.MIN_VALUE;
18         for (int i = 0; i < 24 * 60; i++) {
19             if (mark[i]) {
20                 if (first != Integer.MAX_VALUE) {
21                     min = Math.min(min, i - prev);
22                 }
23                 first = Math.min(first, i);
24                 last = Math.max(last, i);
25                 prev = i;
26             }
27         }
28         min = Math.min(min, (24 * 60 - last + first));
29         return min;
30     }
31 }

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