Code & Fun

第46天。

今天的题目是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?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.

  1. 1 step + 1 step
  2. 2 steps
    Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.

  1. 1 step + 1 step + 1 step
  2. 1 step + 2 steps
  3. 2 steps + 1 step

首先,要到达第n个台阶,我们需要先到n-1或n-2台阶,只要到达n-1和n-2台阶处,我们就能够通过一步到达第n个台阶,这时可以写出这样的递推式:

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climbStairs(n) = climbStairs(n-1) + climbStairs(n-2);
climbStairs(0) = climbStairs(1) = 1;

熟悉的话,可以一眼看出这是斐波那契数列.

这样的话,我们可以很容易写出:

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int climbStairs(int n) {
if (n == 0 || n == 1) return 1;
return climbStairs(n-1) + climbStairs(n-2);
}

但是这样会出现超时的情况,我们可以用一个数组来记录整个斐波那契数列,然后返回适当的值即可:

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int climbStairs(int n) {
vector<int> vec(n+1,1);
for(int i = 2;i <= n;i++) {
vec[i] = vec[i-1] + vec[i-2];
}
return vec[n];
}

这样的时间复杂度和空间复杂度都是O(n).

我们可以把空间复杂度降到O(1):

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int climbStairs(int n) {
int a = 0,b = 1,t;
while(n--) {
t = a+b;
a = b;
b = t;
}
return b;
}

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