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  • 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?

    分析:考虑走第n步时的情况,可以从第n-1个台阶走一步,也可以从第n-2个台阶走两步。即f(n) = f(n-1) + f(n-2),同时f(1) = 1, f(2) = 2.

    1. 使用递归,结果超时了。

    class Solution {
    public:
        int climbStairs(int n) {
            if(n == 1) return 1;
            if(n == 2) return 2;
            else return(climbStairs(n-1) + climbStairs(n-2));
        }
    };
    View Code

    2. 和斐波那契亚数列差不多,于是用了for循环来代替,2ms。

     1 class Solution {
     2 public:
     3     int climbStairs(int n) {
     4         if(n <= 2) return n;
     5         
     6         int *result = new int[n];
     7         result[0] = 1;
     8         result[1] = 2;
     9         for(int i = 2; i < n; i++)
    10             result[i] = result[i-1] + result[i-2];
    11             
    12         return result[n-1];
    13     }
    14 };
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/4442650.html
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