Leetcode刷题记——70. Climbing Stairs（爬楼梯）

一、题目叙述：

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.

Subscribe to see which companies asked this question.

给定台阶总数n，每次只能爬1个楼梯或两个楼梯，问总共有多少种不同的爬法。

二、解题思路：

Esay题，但我写了好一会，简单来讲，直接用数学公式计算阶乘等等即可，但注意，这里要注意int数值越界问题。最终把临时变量sum1换成double类型的才通过。（肯定有好的方法啦。。。）

（1）求共有多少种爬楼梯方法，爬几次2，爬几次1。

（2）针对上面每种情况，计算不同的组合，最后将每种情况下计算出的组合数相加即为结果，注意存放阶乘的临时变量若为int型有可能越界，所以此处的处理方法是将其用double类型存储了。。。

三、源码：

public class Solution  
{  
	 public int climbStairs(int n) 
	 {
		
		int sum = 1; //sum of distinct ways
	    int twosum = n / 2;
	    for (int i = 1; i <= twosum; i++)
	    {
	    	double sum1 = 1;
	    	int q;
	    	int onesum = n - 2 * i;
	    	if (onesum > i)  
	    	{
	    		q = onesum;
	    		for (int j = 1; j <= i; j++)
	    		{
	    			sum1= sum1 * (1 + q);
	    			q ++;
	    		}
	    		for (int k = i; k > 1; k--)
	    			sum1 = sum1 / k;
	    		sum += sum1;
	    	}
	    	else
	    	{
	    		q = i;
	    		for (int j = 1; j <= onesum; j++)
	    		{
	    			sum1= sum1 * (1 + q);
	    			q ++;
	    		}
	    		for (int k = onesum; k > 1; k--)
	    			sum1 = sum1 / k;
	    		sum += sum1;
	    	}
	    		
	    		
	    }
	    return sum;
	    
	 }
    public static void main(String args[])      
    {      
              
    	//String a = "";
    	//String b = "";
      //  int[] digits = {0};      
        Solution solution = new Solution();      
     //   int[][] abc = {{2}};  
      //  int[] b = {2,3,4};  
       // for(int i = 0; i < abc.length; i ++)      
     //   solution.rotate(abc);
         System.out.print(solution.climbStairs(35));      
          
    }      
}