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?

import java.math.BigInteger;

public class Solution {
    public int climbStairs(int n) {
        int num2s = n / 2;
        int ways = 0;
        int num1s = n - num2s * 2;
        for(int i = num2s; i >= 0; i--){
            ways += selectMFromN(num2s + num1s, num2s);
            num2s--;
            num1s += 2;
        }
        return ways;
    }
    /**
     * 计算C(N,M) = n * (n - 1) *...(n - m + 1) / m!
     * @param n
     * @param m
     * @return
     */
    public int selectMFromN(int n, int m){
        BigInteger fenzi = new BigInteger("1");
        BigInteger fenmu = new BigInteger("1");
        for(int i = n ; i >= n - m + 1;i--){
            fenzi = fenzi.multiply(new BigInteger(String.valueOf(i)));
        }
        for(int i = m ; i >= 1;i--){
            fenmu = fenmu.multiply(new BigInteger(String.valueOf(i)));
        }
        return (fenzi.divide(fenmu)).intValue();
    }
}

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

public class Solution {
    public int climbStairs(int n) {
        if(0 == n)
            return 0;
        if(1 == n)
            return 1;
        int pre = 1;
        int current = 1;
        
        for(int i = 2; i <= n; i++){
            int temp = pre + current;
            pre = current;
            current = temp;
        }
        
        return current;
    }
    
}

 用DP确实计算量要小，效率更高

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