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

//这道题其实就是斐波那契数列，用递归会超时
//Approach: Recursive Time: O(n!), Space: O(1)  Tim Limited Exceed
public int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }
        
        return climbStairs(n - 1) + climbStairs(n - 2);
    }

//Approach: Iteration Time: O(n)
//1. Your runtime beats 2.35 % of java submissions.   
 public int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }
        
        int first = 1;
        int second = 2;
        int result = 0;
        
        for (int i = 3; i <= n; i++) {
            result = first + second;
            first = second;
            second = result;
        }
        
        return result;
    }
//2. Your runtime beats 48.13 % of java submissions.
    public int climbStairs(int n) {
        if (n <= 1) return 1;
        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];
    }

//3. Your runtime beats 100.00 % of java submissions
    public int climbStairs(int n) {
        int a = 1, b = 1;
        while (n > 0) {
            b += a; 
            a = b - a;
            n--;
        }
        return a;
    }