Description
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.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
Explain
这道题是关于动态规划的。很经典。这道题如果用递归做，会出现太慢的情况。虽然递归的思想我个人觉得是很正确的。那没办法，只能用动态规划做。每次只能走一阶或两阶。那么我们想想，如果要到第三阶，这个时候就相当于已经走了一阶，现在选择两阶和已经走了两阶，现在选择走一阶。那么 DP[n] = DP[n-1] + DP[n-2] 动态规划的状态方程是不是就出来了。那么下面上代码
Code
class Solution {
public:
int climbStairs(int n) {
int a, b, c, temp;
a = 1;
b = 2;
if (n == 1) return 1;
for (int i = 3; i <= n; i++) {
temp = b;
b = b + a;
a = temp;
}
return b;
}
};
