【leetcode】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?

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题解，变相的斐波那契数列。因为最后一步到达n的时候有两种选择，走1步走到台阶n和走2步走到台阶n。如果走一步走到台阶n，那么之前的n-1级台阶有f(n-1)种走法；如果走2步走到台阶n，那么之前的n-2级台阶有f(n-2)种走法，所以一共有f(n-1)+f(n-2)步走法。

代码如下：

 1 class Solution {
 2 public:
 3     int climbStairs(int n) {
 4         vector<int> num_method(n+1);
 5         num_method[0] = num_method[1] = 1;
 6         for(int i = 2;i <= n;i ++)
 7             num_method[i] = num_method[i-1]+num_method[i-2];
 8         return num_method[n];
 9     }
10 };

 Java代码:不需要数组，只需要保存前面两级台阶的走法数即可。

public class Solution {
    public int climbStairs(int n) {
        if(n == 1)
            return 1;
        if(n == 2)
            return 2;
        int first = 1;
        int second = 2;
        int answer = 0;
        for(int i = 3;i <= n;i++)
        {
            answer = first + second;
            first = second;
            second = answer;
        }
        return answer;
    }
}