Given an integer array?nums?and an integer?k, modify the array in the following way:

• choose an index?i?and replace?nums[i]?with?-nums[i].

You should apply this process exactly?k?times. You may choose the same index?i?multiple times.

Return?the largest possible sum of the array after modifying it in this way.

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Example 1:

Input: nums = [4,2,3], k = 1Output: 5Explanation: Choose index 1 and nums becomes [4,-2,3].

Example 2:

Input: nums = [3,-1,0,2], k = 3Output: 6Explanation: Choose indices (1, 2, 2) and nums becomes [3,1,0,2].

Example 3:

Input: nums = [2,-3,-1,5,-4], k = 2Output: 13Explanation: Choose indices (1, 4) and nums becomes [2,3,-1,5,4].

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Constraints:

• 1 <= nums.length <= 104

• -100 <= nums[i] <= 100

• 1 <= k <= 104

  Easy 題目，如果是小于0的，就取絕對值，如果等于0，說明前面小于0的就已經(jīng)遍歷完了，就可以直接求和了。

• 但是如果所有的都遍歷完了，k還是大于0的話，那么就只能對數(shù)組排序一下，然后對第一個值取反。一直取。即可。

• 然后求和；

Runtime:?2 ms, faster than?98.19%?of?Java?online submissions for?Maximize Sum Of Array After K Negations.

Memory Usage:?41.8 MB, less than?62.99%?of?Java?online submissions for?Maximize Sum Of Array After K Negations.