剑指 Offer II 035. 最小时间差

1. 将字符串转换为以分钟为单位的数组。
2. 对数组进行排序。

3. 两两比较。注意，尾部需要经过 24 * 60 格式化一下，即 time = 24 * 60 - time[n - 1]

   class Solution {
    public int findMinDifference(List<String> timePoints) {
        // 排序
        if (timePoints == null || timePoints.size() == 0 || timePoints.size() > 60 * 24) return 0;
        int n = timePoints.size();
        int[] times = new int[n];
   
        for (int i = 0; i < n; i++) {
            String[] timeStr = timePoints.get(i).split(":");
            // 转换成分
            times[i] = Integer.parseInt(timeStr[0]) * 60 + Integer.parseInt(timeStr[1]);
        }
   
        // 排序
        Arrays.sort(times);
   
        int min = Integer.MAX_VALUE;
        for (int i = 1; i < n; i++) {
            min = Math.min(min, times[i] - times[i - 1]);
        }
   
        min = Math.min(min, 24 * 60 - times[n - 1] + times[0]);
   
        return min;
    }
   }