Climbing Stairs

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?

 1 import java.math.BigInteger;
 2 
 3 public class Solution {
 4     public int climbStairs(int n) {
 5         int num2s = n / 2;
 6         int ways = 0;
 7         int num1s = n - num2s * 2;
 8         for(int i = num2s; i >= 0; i--){
 9             ways += selectMFromN(num2s + num1s, num2s);
10             num2s--;
11             num1s += 2;
12         }
13         return ways;
14     }
15     /**
16      * 计算C(N,M) = n * (n - 1) *...(n - m + 1) / m!
17      * @param n
18      * @param m
19      * @return
20      */
21     public int selectMFromN(int n, int m){
22         BigInteger fenzi = new BigInteger("1");
23         BigInteger fenmu = new BigInteger("1");
24         for(int i = n ; i >= n - m + 1;i--){
25             fenzi = fenzi.multiply(new BigInteger(String.valueOf(i)));
26         }
27         for(int i = m ; i >= 1;i--){
28             fenmu = fenmu.multiply(new BigInteger(String.valueOf(i)));
29         }
30         return (fenzi.divide(fenmu)).intValue();
31     }
32 }

这里要求C(N,M)有溢出，我用BigInteger处理的

ps:用斐波拉契数列实现，或者动态规划DP

 1 public class Solution {
 2     public int climbStairs(int n) {
 3         if(0 == n)
 4             return 0;
 5         if(1 == n)
 6             return 1;
 7         int pre = 1;
 8         int current = 1;
 9         
10         for(int i = 2; i <= n; i++){
11             int temp = pre + current;
12             pre = current;
13             current = temp;
14         }
15         
16         return current;
17     }
18     
19 }

 用DP确实计算量要小，效率更高

 1 public class Solution {
 2     public int climbStairs(int n) {
 3         int steps[] = new int[n + 1];
 4         for(int i = 0; i <= n; i++){
 5             if(0 == i || 1 == i)
 6                 steps[i] = 1;
 7             else{
 8                 steps[i] = steps[i - 1] + steps[i - 2];
 9             }
10         }
11         return steps[n];
12     }
13 }

 

posted on 2014-11-12 15:38  luckygxf  阅读(160)  评论(0)  编辑  收藏  举报

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