Climbing Stairs
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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?
刚开始看到题目简单的写了个递归,提交超时。后来又仔细推演,发现这不就是斐波那契数列么,于是简单的写个循环提交AC
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
刚开始看到题目简单的写了个递归,提交超时。后来又仔细推演,发现这不就是斐波那契数列么,于是简单的写个循环提交AC
int climbStairs(int n) {//if (n < 1) return 0;//else if (n == 1) return 1;//else if (n == 2) return 2;//else return climbStairs(n-1) + climbStairs(n-2); int first = 0, second = 1;int fib = first + second;for (int i = 2; i <= n; i++){first = second;second = fib;fib = first + second;}return fib;}
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- Climbing Stairs
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