LeetCode Minimum Time Difference
原题链接在这里:https://leetcode.com/problems/minimum-time-difference/description/
题目:
Given a list of 24-hour clock time points in "Hour:Minutes" format, find the minimum minutes difference between any two time points in the list.
Example 1:
Input: ["23:59","00:00"]
Output: 1
Note:
- The number of time points in the given list is at least 2 and won't exceed 20000.
- The input time is legal and ranges from 00:00 to 23:59.
题解:
利用bucket sort. 把时间转化成的位置标记成true. 若是之前已经标记了说明有duplicate, 可以直接return 0.
从头到尾iterate buckets维护最小diff. 再和首尾相差的diff比较取出最小值.
Time Complexity: O(timePoints.size() + 24*60).
Space: O(24*60).
AC Java:
class Solution {
public int findMinDifference(List<String> timePoints) {
boolean [] mark = new boolean[24*60];
for(String s : timePoints){
String [] parts = s.split(":");
int time = Integer.valueOf(parts[0]) * 60 + Integer.valueOf(parts[1]); if(mark[time]){
return 0;
} mark[time] = true;
} int res = 24*60;
int pre = -1;
int first = Integer.MAX_VALUE;
int last = Integer.MIN_VALUE;
for(int i = 0; i<24*60; i++){
if(mark[i]){
if(pre != -1){
res = Math.min(res, i-pre);
} pre = i;
first = Math.min(first, i);
last = Math.max(last, i);
}
} res = Math.min(res, first+24*60-last);
return res;
}
}
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