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描述
Given an array nums of integers, we must modify the array in the following way: we choose an i and replace nums[i] with -nums[i], and we repeat this process k times in total. (We may choose the same index i multiple times.)
Return the largest possible sum of the array after modifying it in this way.
Example 1:
Input: nums = [4,2,3], k = 1
Output: 5
Explanation: Choose indices (1,) and nums becomes [4,-2,3].
Example 2:
Input: nums = [3,-1,0,2], k = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and nums becomes [3,1,0,2].
Example 3:
Input: nums = [2,-3,-1,5,-4], k = 2
Output: 13
Explanation: Choose indices (1, 4) and nums becomes [2,3,-1,5,4].
Note:
1 <= nums.length <= 10000
1 <= k <= 10000
-100 <= nums[i] <= 100
解析
根据题意，就是执行 k 次操作，该操作就是必须将 nums 中的某个数变为相反数，经过 k 次操作之后求 nums 的和。直接循环 k 次，每次循环在 nums 里面找出相反数为最大的元素，将其改为其相反数，经过 k 轮循环，最后对 nums 求和即可。
解答
class Solution(object):
def largestSumAfterKNegations(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
while k:
mx = float("-inf")
idx = None
for i,n in enumerate(nums):
if -n>=mx:
mx = -n
idx = i
nums[idx] = -nums[idx]
k-=1
return sum(nums)
运行结果
Runtime: 300 ms, faster than 5.56% of Python online submissions for Maximize Sum Of Array After K Negations.
Memory Usage: 13.5 MB, less than 62.50% of Python online submissions for Maximize Sum Of Array After K Negations.
解析
上面的解法好理解，但是效率不高，所以用数据结构中的堆可以更有效，思路和上面一样，每次从堆 nums 中找出最小值，并将其变为相反数，然后更新当前 nums 的和，重复 k 次即可得到答案。
解答
class Solution(object):
def largestSumAfterKNegations(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
result = sum(nums)
heapq.heapify(nums)
while k:
mn = heapq.heappop(nums)
heapq.heappush(nums, -mn)
result += -mn*2
k -=1
return result
运行结果
Runtime: 44 ms, faster than 34.72% of Python online submissions for Maximize Sum Of Array After K Negations.
Memory Usage: 13.4 MB, less than 62.50% of Python online submissions for Maximize Sum Of Array After K Negations.
原题链接：leetcode.com/problems/ma…
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