leetcode396. Rotate Function

yintaolaowanzi 發布于2019-08-16 13:50 / 554人閱讀

摘要：題目要求代表對數組在位置上進行順時針的旋轉后生成的數組。暴力循環按照題目的要求，執行兩次循環即可以獲得的所有值，只需要從中比較最大值即可。

題目要求

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.

Bk代表對數組A在位置k上進行順時針的旋轉后生成的數組。F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]，要求返回獲得的最大的F(k)的值。

暴力循環

按照題目的要求，執行兩次循環即可以獲得F(k)的所有值，只需要從中比較最大值即可。

    public int maxRotateFunction(int[] A) {
        if(A == null || A.length == 0) return 0;
        int max = Integer.MIN_VALUE;
        for(int i = 0 ; i < A.length ; i++) {
            int value = 0;
            for(int j = 0 ; i < A.length ; j++) {
                value += j * A[(j+i)%A.length];
            }
            max = Math.max(value, max);
        }
        return max;
    }

數學思路

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
F(k-1) = 0 * Bk-1[0] + 1 * Bk-1[1] + ... + (n-1) * Bk-1[n-1] 

F(k) = F(k-1) + sum - n*Bk[0]

k = 0 Bk[0] = A[0]
k = 1 Bk[0] = A[len-1]
k = 2 Bk[0] = A[len-2]
...

    public int maxRotateFunction(int[] A) {
        if(A == null || A.length == 0) return 0;
        int F = 0;
        int sum = 0;
        for(int i = 0 ; i

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