You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
分析，该问题类似于斐波那契序列问题，直观上采用递归来做，f(n)=f(n-1)+f(m-2)，然而时间复杂度太大，因此类似问题可以采用动态规划的思想，如下代码
public class Solution {
public int climbStairs(int n) {
if(n<=2)
return n;
else
{
int[] kinds=new int[n];
kinds[0]=1;
kinds[1]=2;
for(int i=2;i<n;i++)
{
kinds[i]=kinds[i-1]+kinds[i-2];
}
return kinds[n-1];
}
}
}
原文地址：http://7061299.blog.51cto.com/7051299/1640764
