# 949.Largest Time For Given Digits

**949.Largest Time For Given Digits**

难度:Easy

> 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
```

暴力解法，列出四个数字的全排列，然后比较有效的最大值。

```
class Solution {

public:
string largestTimeFromDigits(vector<int>& A) {
		int res = -1;
		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;
					int hours = 10 * A[i] + A[j];
					int mins = 10 * A[k] + A[l];
					if (hours < 24 && mins < 60)
						res = max(res, hours * 60 + mins);
				}
		if (res >= 0) {
			char c[6];
			sprintf(c, "%.2d:%.2d", res / 60, res % 60);
			return c;
		}
		return "";
	}
};
```

> 执行用时 :4 ms, 在所有 C++ 提交中击败了75.93%的用户\
> 内存消耗 :8.1 MB, 在所有 C++ 提交中击败了97.78%的用户


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