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
Solution:
class Solution {
public int climbStairs(int n) {
/*
dp[i] means how many step to current i staircase
2 _
1 _
0 _
2 + 1
1+1
dp[i] 1
*/
if (n == 1){
return 1;
}
int[] dp = new int[n];
dp[0] = 1;
dp[1] = 2;
for (int i = 2; i < n; i++){
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n - 1];
}
}
// TC: O(n)
// SC: O(n)
Brute Force
class Solution {
public int climbStairs(int n) {
return help(0, n);
}
public int help(int i, int n){
if (i > n){
return 0;
}
if (i == n){
return 1;
}
return help(i+1, n) + help(i+2, n);
}
}
//TLE
//TC: O(2^n)
//SC: O(n)
Dynamic:
class Solution {
public int climbStairs(int n) {
if (n == 0 || n == 1 || n == 2 || n == 3){
return n;
}
int[] dp = new int[n + 1];
dp[0] = 0;
dp[1] = 1;
dp[2] = 2;
dp[3] = 3;
for (int i = 4; i < n + 1; i++){
dp[i] = dp[i-1] + dp[i-2];
}
return dp[n];
}
}
// TC: O(n)
// SC: O(n)
/*
n = 4
dp: 0,1,2,3,4
[0,1,2,3, ]
dp[i] = dp[i-1] + dp[i-2]
0
/ \
1 2
/ \ / \
2 3 3 4
/ \ / /
3 4 4 4
/
4
*/
https://leetcode.cn/problems/climbing-stairs/solutions/2560716/jiao-ni-yi-bu-bu-si-kao-dong-tai-gui-hua-7zm1/