70. Climbing Stairs（动态规划 爬台阶，一次只能爬1,2两节）

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?

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

f(1) = 1

f(2) = 2

f(3) = f(n-1)+f(n-2)

f(t) = f(t-1)+ f(t-2)

递归超时：：

class Solution {
    public int climbStairs(int n) {
        if(n<3) return n;
        return climbStairs(n-1) + climbStairs(n-2);
    }
}

动态规划！！！！

class Solution {
    public int climbStairs(int n) {
        int dp[] = new int[n+1];
        for(int i=0;i<=n;i++){
            if(i<3)
            dp[i] = i;
            else
                dp[i] = dp[i-1]+dp[i-2];
        }
        return dp[n];
    }
    
}

Fibonacci Number:

class Solution {
    public int climbStairs(int n) {
        if(n<3) return n;
        int t1 = 1;
        int t2 = 2;
        int t3;
        for(int i=2;i<n;i++){
            t3 = t1+t2;
            t1 = t2;
            t2 = t3;
        }
        return t2;
    }
    
}