变态跳台阶
变态跳台阶
http://www.nowcoder.com/questionTerminal/22243d016f6b47f2a6928b4313c85387
解题思路：有n个台阶时，F(n)=F(n-1)+F(n-2)+F(n-3)+……+F(1)+F(0).又因为F(1)=F(0)=1,经计算可得，F(n)=2^(n-1)
import java.lang.Math;
public class Solution {
public int JumpFloorII(int target) {
if(target<2){
return 1;
}
return (int)Math.pow(2,target-1);
}
}
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