You are given an integer array?nums?and an integer?k. Append?k?unique positive?integers that do?not?appear in?nums?to?nums?such that the resulting total sum is?minimum.

Return?the sum of the?k?integers appended to?nums.

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Example 1:

Input: nums = [1,4,25,10,25], k = 2Output: 5Explanation: The two unique positive integers that do not appear in nums which we append are 2 and 3. The resulting sum of nums is 1 + 4 + 25 + 10 + 25 + 2 + 3 = 70, which is the minimum. The sum of the two integers appended is 2 + 3 = 5, so we return 5.

Example 2:

Input: nums = [5,6], k = 6Output: 25Explanation: The six unique positive integers that do not appear in nums which we append are 1, 2, 3, 4, 7, and 8. The resulting sum of nums is 5 + 6 + 1 + 2 + 3 + 4 + 7 + 8 = 36, which is the minimum. The sum of the six integers appended is 1 + 2 + 3 + 4 + 7 + 8 = 25, so we return 25.

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Constraints:

• 1 <= nums.length <= 10(5)

• 1 <= nums[i] <= 10(9)

• 1 <= k <= 10(8)

第一個(gè)方法就是常規(guī)思路，但是TLE(TIME LIMIT EXCEED)超時(shí)了，

所以 只能轉(zhuǎn)換思路，因?yàn)镵會(huì)很大，為了減少遍歷k內(nèi)部的數(shù)字，于是先把k以內(nèi)的數(shù)字求和，

然后去判斷數(shù)字在數(shù)組中是否存在（為了方便，把數(shù)組放在set-集合中），如果存在了，則sum-i, cnt++（這里cnt就要去用大于k的數(shù)字去添加了）。

遍歷后再去根據(jù)cnt的大小去遍歷大于k的數(shù)字，這時(shí)候還是要判斷在不在set中，

遍歷完就可以return了。

Runtime:?37 ms, faster than?60.47%?of?Java?online submissions for?Append K Integers With Minimal Sum.

Memory Usage:?59.2 MB, less than?75.35%?of?Java?online submissions for?Append K Integers With Minimal Sum.