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climbing-stairs

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Climbing Stairs – LeetCode 70

Problem

Description

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

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

Answer

Original

Code

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class Solution {
public:
	int climb(int i, int n, int memo[]){
		if(i > n) return 0;
		if(i == n) return 1;
		if(memo[i] > 0) return memo[i];
		memo[i] = climb(i+1,n,memo) + climb(i+2,n,memo);
		return memo[i];
	}

	int climbStairs(int n){
		int memo[n+1] = {0};
		return climb(0,n,memo);
    }
};

思路

用递归法模拟步进，妙在左侧走1步，右侧走2步。同时使用memo来缓存先行求值的答案。时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$3$ ms，排名$98.89\%$

Better

思路

用动态规划，时间复杂度$O(n)$，空间复杂度$O(n)$。更多思路
耗时$0$ ms，排名$71.58\%$

Code

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class Solution {
public:
	int climbStairs(int n){
		if(n == 1) return 1;
        unsigned pp = 1, prev = 2,curr;
        for(unsigned i = 2; i != n; ++i){
            prev += pp;
            pp = prev - pp;
        }
        return prev;
    }
};