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?
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
f(1) = 1
f(2) = 2
f(3) = f(n-1)+f(n-2)
f(t) = f(t-1)+ f(t-2)
递归超时：：
class Solution {
public int climbStairs(int n) {
if(n<3) return n;
return climbStairs(n-1) + climbStairs(n-2);
}
}
动态规划！！！！
class Solution {
public int climbStairs(int n) {
int dp[] = new int[n+1];
for(int i=0;i<=n;i++){
if(i<3)
dp[i] = i;
else
dp[i] = dp[i-1]+dp[i-2];
}
return dp[n];
}
}
Fibonacci Number:
class Solution {
public int climbStairs(int n) {
if(n<3) return n;
int t1 = 1;
int t2 = 2;
int t3;
for(int i=2;i<n;i++){
t3 = t1+t2;
t1 = t2;
t2 = t3;
}
return t2;
}
}