Leetcode-Climbing Stairs

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;
    }
};