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.
给出一个整型数组A,设n为其长度。 假设Bk是将A进行k此顺时针旋转后的数组,我们定义一个A的“旋转函数”F,如下: F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]. 注意: n保证不会超过10的5次方 例子: 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 所以 F(0), F(1), F(2), F(3) 中最大的值为 F(3) = 26.
这个题目的意思就是对A进行旋转求多项式,每次旋转系数移动一次,旋转一次求出一个结果,看哪个最大就返回哪个。
我的做法是直接进行每一个的计算然后找最大的,代码挺简单,时间复杂度是O(n平方),很长。
public class Solution {
public int maxRotateFunction(int[] A) {
if (A.length == 0) return 0;
int result = -2147483648;
for (int i = 0; i < A.length; i++) {
int sum = sum(A, A.length - i);
if (sum > result) result = sum;
}
return result;
}
public int sum(int[] A, int index) {
int sum = 0;
for (int i = 0; i < A.length; i++) {
if (index >= A.length) index = 0;
sum += i * A[index];
index ++;
}
return sum;
}
}
public class Solution {
public int maxRotateFunction(int[] A) {
int allSum = 0;
int len = A.length;
int F = 0;
for (int i = 0; i < len; i++) {
F += i * A[i];
allSum += A[i];
}
int max = F;
for (int i = len - 1; i >= 1; i--) {
F = F + allSum - len * A[i];
max = Math.max(F, max);
}
return max;
}
}
这个做法的好处在于只需要O(n)的时间,快很多。他对题目的要求进行了一些数学计算,然后得出了一个方便计算的式子,过程如下:
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] = 0 * Bk[1] + 1 * Bk[2] + ... + (n-2) * Bk[n-1] + (n-1) * Bk[0]
那么,
F(k) - F(k-1) = Bk[1] + Bk[2] + ... + Bk[n-1] + (1-n)Bk[0] = (Bk[0] + ... + Bk[n-1]) - nBk[0] = sum - nBk[0]
因此,
F(k) = F(k-1) + sum - nBk[0]
那Bk[0]是什么呢?
k = 0; B[0] = A[0]; k = 1; B[0] = A[len-1]; k = 2; B[0] = A[len-2]; ...
这样,也就有了上面的代码了。