算法设计Week8 LeetCode Algorithms Problem #70 Climbing Stairs

题目描述:

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?

Note: Given n will be a positive integer.

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题目分析:

使用动态规划算法，设climb[i]为爬上第i级台阶不同方法的总数，那么有两种方式可以爬上第i级台阶：
1. 爬上第i-1级台阶后再向上爬一级台阶；
2. 爬上第i-2级台阶后再向上爬两级台阶。
这样，就可以得到递推式：
climb[i] = climb[i - 1] + climb[i - 2].
此外，climb[1] = 1, climb[2] = 2。
具体代码如下（注意代码中climb数组下标从0开始），算法的时间复杂度为O(n)，空间复杂度为O(n)。

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