70. Climbing Stairs

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

Solution1:递归

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

但是运行显示 Time limited exceed.

网上大佬说是因为递归的效率太低，会产生很多分支，所以放弃使用递归。

Solution2:

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

这个方法使用了Dynamic Programming提高效率。来自Grandyang大佬。

solution3

“我们可以对空间进行进一步优化，我们只用两个整型变量a和b来存储过程值，首先将a+b的值赋给b，然后a赋值为原来的b，所以应该赋值为b-a即可。这样就模拟了上面累加的过程，而不用存储所有的值，参见代码如下：”

我去这是什么神仙方法，

public class Solution {
    public int climbStairs(int n) {
        int a = 1, b = 1;
        while (n-- > 0) {
            b += a; 
            a = b - a;
        }
        return a;
    }
}

 附递归和动态规划算法题目详解：

http://www.cnblogs.com/DarrenChan/p/8734203.html#_label1

3. 迭代，和求Fibonacci数列第n项一样，每次把prev和cur向右移一位

class Solution {
 public int climbStairs(int n) {
     int cur = 1;
     int prev = 0;
     for(int i = 1; i <= n; i++){
         int tmp = cur;
         cur += prev;
         prev = tmp;
     }
     return cur;
    }
}