There are two mice and?n?different types of cheese, each type of cheese should be eaten by exactly one mouse.

A point of the cheese with index?i?(0-indexed) is:

• reward1[i]?if the first mouse eats it.

• reward2[i]?if the second mouse eats it.

You are given a positive integer array?reward1, a positive integer array?reward2, and a non-negative integer?k.

Return?the maximum?points the mice can achieve if the first mouse eats exactly?k?types of cheese.

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Example 1:

Input: reward1 = [1,1,3,4], reward2 = [4,4,1,1], k = 2

Output: 15

Explanation: In this example, the first mouse eats the 2nd?(0-indexed) and the 3rd?types of cheese, and the second mouse eats the 0th?and the 1st types of cheese. The total points are 4 + 4 + 3 + 4 = 15. It can be proven that 15 is the maximum total points that the mice can achieve.

Example 2:

Input: reward1 = [1,1], reward2 = [1,1], k = 2

Output: 2

Explanation: In this example, the first mouse eats the 0th?(0-indexed) and 1st?types of cheese, and the second mouse does not eat any cheese. The total points are 1 + 1 = 2. It can be proven that 2 is the maximum total points that the mice can achieve.

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Constraints:

• 1 <= n == reward1.length == reward2.length <= 105

• 1 <= reward1[i],?reward2[i] <= 1000

• 0 <= k <= n

k個(gè)reward1+n-k個(gè)reward2 的最大值，如何取判斷，先將reward1-reward2，排序求出前k個(gè)最大值，然后再加reward2的數(shù)組和，這樣就是k個(gè)reward1 +n-k個(gè)reward2的最大值 了。

Runtime:?13 ms, faster than?100.00%?of?Java?online submissions for?Mice and Cheese.

Memory Usage:?50.5 MB, less than?100.00%?of?Java?online submissions for?Mice and Cheese.