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Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[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: A = [4,2,3], K = 1Output: 5Explanation: Choose indices (1,) and A becomes [4,-2,3].

Example 2:

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

Example 3:

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

Note:

1. 1 <= A.length <= 10000
2. 1 <= K <= 10000
3. -100 <= A[i] <= 100

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非常简单，居然能wa一次，不应该啊。。。自己想一下，确保脑子里没有bug

class Solution:    def largestSumAfterKNegations(self, A, K):        sa = sorted(A)        i = 0        for i in range(K):            if (sa[i] < 0):                sa[i] = -sa[i]            else:                min_val = min(sa)                res = sum(sa)                delta = min_val                for j in range(K - i):                    res -= 2*delta                    delta = -delta                return res        return sum(sa)s = Solution()print(s.largestSumAfterKNegations([4,2,3], 1))

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