Given an array of 4 digits, return the largest 24 hour time that can be made.
The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight.
Return the answer as a string of length 5. If no valid time can be made, return an empty string.
Example 1:
Input: [1,2,3,4]
Output: "23:41"
Example 2:
Input: [5,5,5,5]
Output: ""
Note:
A.length == 4
0 <= A[i] <= 9
class Solution { public String largestTimeFromDigits(int[] A) { String ans = ""; for (int i = 0; i < 4; ++i) { for (int j = 0; j < 4; ++j) { for (int k = 0; k < 4; ++k) { for(int l = 0; l < 4; l++) { if (i == j || i == k || i == l || j == k || j == l || k == l) continue; // avoid duplicate among i, j & k. String h = "" + A[i] + A[j], m = "" + A[k] + A[l], t = h + ":" + m; // hour, minutes, & time. if (h.compareTo("24") < 0 && m.compareTo("60") < 0 && ans.compareTo(t) < 0) ans = t; // hour < 24; minute < 60; update result. } } } } return ans; } }
垃圾题。。permutation数组里的四位,符合条件且更大就update