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70. Climbing Stairs

You are climbing a staircase. It takes n steps to reach the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Example 1:

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

  1. 1 step + 1 step
  2. 2 steps

Example 2:

Input: n = 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

Constraints:

1 <= n <= 45

跳到某一级有2种方式，从它上一级跳上来，或从它上上级跳上来，所以 f(n)=f(n-1)+f(n-2) 。会超时的递归算法如下，由于涉及到大量的重复运算，时间复杂度 O(an)。

class Solution {
    public int climbStairs(int n) {
        if(n<=3) return n;
        return climbStairs(n-1)+climbStairs(n-2);
    }
}

这题的结果是斐波那契数列，每次只需用到前面2个数字，可以将前2个算出的结果存起来，减少重复计算。时间复杂度 O(n)

class Solution {
    public int climbStairs(int n) {
        if(n<=2) return n;
        int pre1=1,pre2=2;
        int ret=0;
        for(int i=2;i<n;++i){
            ret=pre1+pre2;
            pre1=pre2;
            pre2=ret;
        }
        return ret;
    }
}