https://leetcode.com/problems/largest-time-for-given-digits/
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 == 40 <= A[i] <= 9
Try all possible times, and remember the largest one.
Algorithm (Java)
Iterate over all permutations
(i, j, k, l) of (0, 1, 2, 3). For each permutation, we can try the time A[i]A[j] : A[k]A[l].
This is a valid time if and only if the number of hours
10*A[i] + A[j] is less than 24; and the number of minutes 10*A[k] + A[l] is less than 60.
We will output the largest valid time.
// Solution inspired by @rock class Solution { public String largestTimeFromDigits(int[] A) { int ans = -1; // Choose different indices i, j, k, l as a permutation of 0, 1, 2, 3 for (int i = 0; i < 4; ++i) for (int j = 0; j < 4; ++j) if (j != i) for (int k = 0; k < 4; ++k) if (k != i && k != j) { int l = 6 - i - j - k; // For each permutation of A[i], read out the time and // record the largest legal time. int hours = 10 * A[i] + A[j]; int mins = 10 * A[k] + A[l]; if (hours < 24 && mins < 60) ans = Math.max(ans, hours * 60 + mins); } return ans >= 0 ? String.format("%02d:%02d", ans / 60, ans % 60) : ""; } }
The inner most loop at most iterates 4 * 4 * 4 = 64 times.
A[i], A[j], A[k], & A[l] are the 4 elements of A, where i, j, k & l are the permutation of 0, 1, 2, & 3. Therefore, since i + j + k + l = 0 + 1 + 2 + 3 = 6, we have l = 6 - i - j - k.
A[i], A[j], A[k], & A[l] are the 4 elements of A, where i, j, k & l are the permutation of 0, 1, 2, & 3. Therefore, since i + j + k + l = 0 + 1 + 2 + 3 = 6, we have l = 6 - i - j - k.
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) {
if (i == j || i == k || j == k) continue; // avoid duplicate among i, j & k.
String h = "" + A[i] + A[j], m = "" + A[k] + A[6 - i - j - k], 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;
}public String largestTimeFromDigits(int[] a) {
LinkedList<String> q = new LinkedList<>();
q.add("");
for (int n : a)
for (int size = q.size(); size > 0; size--) {
String s = q.poll();
for (int i = 0; i <= s.length(); i++)
q.add(s.substring(0, i) + n + s.substring(i));
}
String largest = "";
for (String s : q) {
s = s.substring(0, 2) + ":" + s.substring(2);
if (s.charAt(3) < '6' && s.compareTo("24:00") < 0 && s.compareTo(largest) > 0)
largest = s;
}
return largest;
} /**
* 4个数字全排列总共有24种可能，判断每一种可能是否能组成合法时间值，如果能，再和当前保存的最大值进行比较；
* 最大值是一个int值，用来表示分钟数；
*/
public String largestTimeFromDigits(int[] array) {
int largestTime = -1;
// 暴力枚举出每种可能
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
if (j != i) {
for (int k = 0; k < 4; k++) {
if (k != i && k != j) {
// 0,1,2,3 总和为6，故剩下的index为6-
int l = 6 - i - j - k;
int result = largestTimeHelper(array[i], array[j], array[k], array[l]);
largestTime = Math.max(result, largestTime);
}
}
}
}
}
if (largestTime==-1){
return "";
}
return String.format("%2d:%2d",largestTime/60,largestTime%60);
}
/**
* 判断输入的四个数字按照输入顺序组成的时间是否合法，如果合法，返回分钟数；
*/
public int largestTimeHelper(int a, int b, int c, int d) {
int hours = a * 10 + b;
int min = c * 10 + d;
if (hours < 24 && min < 60) {
// 返回分钟数
return hours * 60 + min;
}
return -1;
}https://www.acwing.com/solution/LeetCode/content/680/
给定一个由 4 位数字组成的数组，返回可以设置的符合 24 小时制的最大时间。
最小的 24 小时制时间是 00:00，而最大的是 23:59。从 00:00 （午夜）开始算起，过得越久，时间越大。
以长度为 5 的字符串返回答案。如果不能确定有效时间，则返回空字符串。
样例
输入：[1,2,3,4]
输出："23:41"
输入：[5,5,5,5]
输出：""
注意
A.length == 40 <= A[i] <= 9
算法
(暴力枚举)
- 枚举所有可能的数字排列，然后判断哪些是合法的时间，最后输出最大的时间即可。
时间复杂度
- 枚举排列的时间复杂度，为 4!。
