题目要求
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?
做这道题容易陷入的思维就是:所有step加起来的总和是n。俺这种思维去解决这个问题的话会显得很复杂。
按动态规划的思维,要爬到第n阶,可以从第n-1阶爬1阶到达,也可以从第n-2阶爬2阶到达,因此爬到第n阶的方法有这么多种:
dp[n] = dp[n - 1] + dp[n - 2]
按此思维的程序代码如下:
1 class Solution { 2 public: 3 int climbStairs(int n) { 4 if(n == 0 || n == 1 || n == 2) 5 return n; 6 7 int * dp = new int[n + 1]; 8 dp[0] = 0; 9 dp[1] = 1; 10 dp[2] = 2; 11 12 for(int i = 3; i < n + 1; i++) 13 dp[i] = dp[i - 1] + dp[i - 2]; 14 15 return dp[n]; 16 } 17 };
时间: 2024-10-08 21:33:08