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?
解题思路：
动态规划DP(Dynamic Programming)入门题。
state: dp[i] 表示爬到第i个楼梯的所有方法的和
function： dp[i] = dp[i-1] + dp[i-2] //因为每次走一步或者两步， 所以f[i]的方法就是它一步前和两步前方法加和
initial： dp[0] = 0; dp[1] = 1
end : return f[n]
Java: Method 1: Time: O(n), Space: O(n)
public int climbStairs(int n) {
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
Java: Method 2: Time: O(n), Space: O(1)
public int climbStairs(int n) {
if (n == 0 || n == 1 || n == 2){
return n;
}
int [] dp = new int[3];
dp[1] = 1;
dp[2] = 2;
for (int i =3; i <= n; i++) {
dp[i%3] = dp[(i-1)%3] + dp[(i-2)%3];
}
return dp[n%3];
}
Java: Method 3: Time: O(n), Space: O(1)
public class Solution {
public int climbStairs(int n) {
int[] dp = new int[]{0,1,2};
if(n < 3) return dp[n];
for(int i = 2; i < n; i++){
dp[0] = dp[1];
dp[1] = dp[2];
dp[2] = dp[0] + dp[1];
}
return dp[2];
}
}
Python:
class Solution:
# DP Time: O(n) Space: O(1)
def climbStairs(self, n):
prev, current = 0, 1
for i in xrange(n):
prev, current = current, prev + current,
return current
# Recursion Time: O(2^n) Space: O(n)
def climbStairs1(self, n):
if n == 1:
return 1
if n == 2:
return 2
return self.climbStairs(n - 1) + self.climbStairs(n - 2)
