0949. Largest Time for Given Digits (E)

Largest Time for Given Digits (E)

题目

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:

1. A.length == 4
2. 0 <= A[i] <= 9

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题意

用给定的4个数字组成最大的24时制的时间。

思路

可以直接回溯法解决。每一位上都有一个值的范围，从最大值往前遍历寻找可能的组合。

或者直接对小时23-0和对分钟59-0进行二重循环遍历。

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代码实现

Java

回溯法

class Solution {
    public String largestTimeFromDigits(int[] A) {
        int[] digits = new int[10];
        for (int num : A) {
            digits[num]++;
        }
        return generate(digits, 0, "");
    }

    private String generate(int[] digits, int index, String s) {
        if (index == 5) {
            return s;
        }

        if (index == 2) {
            return generate(digits, index + 1, s + ":");
        }

        int max = index == 0 ? 2 : index == 1 ? (s.charAt(0) == '2' ? 3 : 9) : index == 3 ? 5 : 9;
        for (int i = max; i >= 0; i--) {
            if (digits[i] > 0) {
                digits[i]--;
                String t = generate(digits, index + 1, s + i);
                if (!t.isEmpty()) {
                    return t;
                } else {
                    digits[i]++;
                }
            }
        }

        return "";
    }
}

二重循环

class Solution {
    public String largestTimeFromDigits(int[] A) {
        Arrays.sort(A);
        
        for (int i = 23; i >= 0; i--) {
            for (int j = 59; j >= 0; j--) {
                int a = i < 10 ? 0 : i / 10;
                int b = i % 10;
                int c = j < 10 ? 0 : j / 10;
                int d = j % 10;
                int[] B = { a, b, c, d };
                Arrays.sort(B);
                boolean valid = true;
                for (int k = 0; k < 4; k++) {
                    if (A[k] != B[k]) {
                        valid = false;
                        break;
                    }
                }
                if (valid) {
                    return "" + a + b + ":" + c + d;
                }
            }
        }

        return "";
    }
}