LeetCode 539. Minimum Time Difference

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题目要求

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:

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.

题意解析

计算最短的时间间隔，注意24小时制值得变化。

解法分析

将时间进行转换，转换成整型，转换方式为hour*24+minute，然后排序，最后计算就可以了。

解题代码

	public int findMinDifference(List<String> timePoints) {
	int[] timeRevert = new int[timePoints.size()];

	for (int i = 0; i < timePoints.size(); i++) {
		String timePoint = timePoints.get(i);
		String[] timeParts = timePoint.split(":");
		int hour = Integer.valueOf(timeParts[0].trim());
		int minute = Integer.valueOf(timeParts[1].trim());
		int revert = hour * 60 + minute;
		timeRevert[i] = revert;
	}
	Arrays.sort(timeRevert);
	int min = Integer.MAX_VALUE;
	for (int i = 0; i < timeRevert.length; i++) {
		if (i < timeRevert.length - 1) {
			if (timeRevert[i + 1] - timeRevert[i] < min) {
				min = timeRevert[i + 1] - timeRevert[i];
			}
		} else {
			if (timeRevert[0] + 24 * 60 - timeRevert[i] < min) {
				min = timeRevert[0] + 24 * 60 - timeRevert[i];
			}
		}
	}
	return min;
}