[LeetCode]70、 Climbing Stairs

题目描述：

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

思路：

public class Solution70 {
    public int climbStairs(int n){
//        //递归的方法，等同于斐波那契数列
//        if (n == 1) return 1;
//        if (n == 2) return 2;
//        return climbStairs(n-1)+climbStairs(n-2);
        
        //动态规划法1
//        if (n == 1) return 1;
//        if (n == 2) return 2;
//        int f1 = 1;
//        int f2 = 2;
//        for(int i = 3; i <= n; i++){
//            int tmp = f1 + f2;
//            f1 = f2;
//            f2 = tmp;
//        }
//        return f2;
        //动态规划法2 动态规划填表法
        int[] res = new int[n+1];
        res[0] = 1;
        res[1] = 1;
        for(int i = 2; i <= n; i++){
             res[i] = res[i-1] + res[i-2]; 
        }
        return res[n];
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution70 solution70 = new Solution70();
        int n = solution70.climbStairs(1);
        System.out.println(n);
    }

}