题目链接: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?

这道题的要求是爬n阶楼梯,每次只可以爬1步或2步,总共有多少种不同方法能爬到顶?

动态规划,假设要爬到第i阶,可以由i-1和i-2阶1次过去,因此dp[i] = dp[i-1] + dp[i-2]。

可以注意到,爬到n阶台阶的方案数目其实就是求斐波那契数列的第n+1位。

时间复杂度:O(n)

空间复杂度:O(n)

 1 class Solution 
 2 {
 3 public:
 4     int climbStairs(int n) 
 5     {
 6         vector<int> v(n + 1);
 7         v[0] = v[1] = 1;
 8         for(int i = 2; i <= n; ++ i)
 9             v[i] = v[i - 1] + v[i - 2];
10         return v[n];
11     }
12 };

由于动态规划的时候只用到了前面2步,因此可以用两个变量记录一下,这样就将空间复杂度降到O(1)了。

时间复杂度:O(n)

空间复杂度:O(1)

 1 class Solution
 2 {
 3 public:
 4     int climbStairs(int n)
 5     {
 6         int n2 = 0, n1 = 1, res = 0;
 7         for(int i = 0; i < n; ++ i)
 8         {
 9             res = n2 + n1;
10             n2 = n1;
11             n1 = res;
12         }
13         return res;
14     }
15 };