Leetcode-70 Climbing Stairs

标签：div   each   nbsp   动态   tair   ++   step   cli   any   

#70.    Climbing Stairs

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?

题解：这道题属于动态规划的题，类似于斐波那契数列，所以采用非递归的方式来解。当楼梯只有一级时，显然只有一种方法,即f(1)=1；当楼梯有两级时，显然有两种方法,即f(2)=2；当楼梯有n级时，f(n) = f(n-1) + f(n-2)；

class Solution {
public:
    int climbStairs(int n) {
        if(n<=2)
        {
            return n;
        }
        int array[n+1]={0};
        array[1]=1;
        array[2]=2;
        for(int i=3;i<=n;i++)
        {
            array[i]=array[i-1]+array[i-2];
        }
        return array[n];
    }
};

Leetcode-70 Climbing Stairs

标签：div   each   nbsp   动态   tair   ++   step   cli   any   

原文地址：http://www.cnblogs.com/fengxw/p/6082860.html