0%

climbing-stairs

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
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

1
2
3
4
5
6
7
8
9
10
11
12
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;
}
};