396. Rotate Function
s0 = 0 * a0 + 1 *a1 + 2 * a2 + 3 * a3
- s1 = 1 * a0 + 2 * a1 + 3 * a2 + 0 * a3
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- a0 - a1 - a2 + 3*a3 = - (a0+a1+a2+a3)+4*a4
所以如果sum表示所有数的和,len表示数组的长度那么每连续两个的差别是:
s0 - s1 = -sum + len*a3
所以公式就是
sn = sn-1 + sum - len * a[len - n]
1 public int maxRotateFunction(int[] A) { 2 int initial = 0; 3 int sum = 0; 4 int len = A.length; 5 for(int i = 0; i < len; i++) { 6 sum += A[i]; 7 initial += i * A[i]; 8 } 9 int max = initial; 10 for(int i = 1; i < len; i++) { 11 initial = initial + sum - len * A[len - i]; 12 max = Math.max(max, initial); 13 } 14 return max; 15 }