45 leetcode - Rotate Function

最新推荐文章于 2025-08-08 21:26:37 发布

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#leetcode #Python #算法 #rotate #function

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本文介绍了一种计算特定旋转函数最大值的方法。该函数应用于整数数组，通过旋转数组并计算每个旋转状态下的函数值来找出最大值。文章提供了一个Python实现示例，展示了如何高效地解决这个问题。

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#!/usr/bin/python
# -*- coding: utf-8 -*-
'''
 Rotate Function
Given an array of integers A and let n to be its length.
Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow:
F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].
Calculate the maximum value of F(0), F(1), ..., F(n-1).
Note:
n is guaranteed to be less than 105.
Example:
A = [4, 3, 2, 6]
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
So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.可以找规律,由F(n)->F(n+1)
'''
class Solution(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        length = len(A)
        if length <= 0:
            return 0

        A_sum = 0
        fun_sum = 0
        max_sum = 0
        for index,val in enumerate(A):
            A_sum += val
            fun_sum += index * val

        max_sum = fun_sum
        while index > 0:
            #利用F(i)->F(i+1),F(i+1) = F(i) - (length - 1) * A[index] + A_sum - A[index]
            fun_sum = fun_sum - (length - 1) * A[index] + A_sum - A[index]
            if fun_sum > max_sum:
                max_sum = fun_sum
            index -= 1

        return max_sum

if __name__ == "__main__":
    s = Solution()
    print s.maxRotateFunction([4,3,2,6])