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Version:
v0.0.0-...-a49be92
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Published: Feb 20, 2024
License: MIT
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README
¶
旋转函数
给定一个长度为 n
的整数数组 nums
。
假设 arrk
是数组 nums
顺时针旋转 k
个位置后的数组,我们定义 nums
的 旋转函数F
为:
F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1]
返回 _F(0), F(1), ..., F(n-1)
中的最大值 _。
生成的测试用例让答案符合 32 位 整数。
示例 1:
输入: nums = [4,3,2,6]
输出: 26
解释:
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
所以 F(0), F(1), F(2), F(3) 中的最大值是 F(3) = 26 。
示例 2:
输入: nums = [100]
输出: 0
提示:
n == nums.length
1 <= n <= 105
-100 <= nums[i] <= 100
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