题目如下 链接
阶梯思路为回溯或DP,类似于Fib数列
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 step + 1 step
- 2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
3. 1 step + 1 step + 1 step
4. 1 step + 2 steps
5. 2 steps + 1 step
递推式:
f
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n
)
=
f
(
n
−
1
)
+
f
(
n
−
2
)
f(n) = f(n-1) + f(n-2)
f(n)=f(n−1)+f(n−2) 一定要注意边界条件
class Solution {
public:
int climbStairs(int n) {
if(n<=2) return n;
int one_step_before = 2;
int two_step_before = 1;
int all_way = 0;
for(int i=2; i<n; i++)
{
all_way = one_step_before + two_step_before; //暂存
two_step_before = one_step_before;
one_step_before = all_way;
}
return all_way;
}
};
class Solution:
def climbStairs(self, n: int) -> int:
'''
:type n: int
:rtype: int
'''
x, y = 1, 1
for _ in range(1, n):
x, y = x+y, x
return x
一个线性的时间复杂度,
public static int climbstairs(int n){
if(n==0 || n==1 || n==2) return n; //!!!
int[] mem = new int[n];
mem[0] = 1;
mem[1] = 2;
for(int i=2; i<n; i++){
mem[i] = mem[i-1] + mem[i-2];
}
return mem[n-1];
}
刚开始没有注意边缘数据。
可以减少变量的个数来节约空间复杂度。
public static int climbstairs(int n){
if(n <=2) return n;
int one_step_before = 2; //i-1
int two_step_before = 1; //i-2
int all_way = 0;
for(int i=2; i<n; i++){
all_way = one_step_before + two_step_before;
two_step_before = one_step_before;
one_step_before = all_way;
}
return all_way;
}