题目:

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?


解题思路:这是一道斐波拉契数列题。看到斐波那契数列就自然想到利用递归,然而在这里递归的效率非常低,荣誉计算的数量增长的非常快,直接导致出现TLE(Time Limit Exceeded)错误。这是因为在递归的时候许多值被重复计算了多次,根据《C和指针》上对斐波那契数列的介绍,在n=30时,Fibonacci(3)的值被计算了317811次。这是非常可怕的,因此我又采用迭代的思路写了一组代码。迭代的思路也比较简单,设定初始f(1)和f(2)的值,在每次迭代中将前两次的计算结果相加作为本次计算结果即可。下面给出递归代码和迭代代码。


递归代码(出现超时错误):

class Solution {
public:
    int climbStairs(int n) {
        int CurrChoices;
        if(n==0){
            CurrChoices=1;
        }else if(n==1){
            CurrChoices=climbStairs(n-1);
        }else{
            CurrChoices=climbStairs(n-1)+climbStairs(n-2);
        }
        return CurrChoices;
    }
};


迭代代码:

class Solution {
public:
    int climbStairs(int n) {
        int CurrChoice,OlderChoice,OldestChoice;
        CurrChoice=OlderChoice=OldestChoice=1;
        if(n==0)return 1;
        if(n==1)return 1;
        while(n>1){
            n-=1;
            OldestChoice=OlderChoice;
            OlderChoice=CurrChoice;
            CurrChoice=OlderChoice+OldestChoice;
        }
        return CurrChoice;
    }
};