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

這道題是關(guān)于動態(tài)規(guī)劃的。很經(jīng)典。這道題如果用遞歸做，會出現(xiàn)太慢的情況。雖然遞歸的思想我個(gè)人覺得是很正確的。那沒辦法，只能用動態(tài)規(guī)劃做。每次只能走一階或兩階。那么我們想想，如果要到第三階，這個(gè)時(shí)候就相當(dāng)于已經(jīng)走了一階，現(xiàn)在選擇兩階和已經(jīng)走了兩階，現(xiàn)在選擇走一階。那么 DP[n] = DP[n-1] + DP[n-2] 動態(tài)規(guī)劃的狀態(tài)方程是不是就出來了。那么下面上代碼

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;
    }
};