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?
Analysis: It's a DP problem. For each n, the last step can either climb 1 or 2 steps. So, f(n) = f(n-1) + f(n-2)
public class Solution {
public int climbStairs(int n) {
// the last step can climb either 1 or 2 steps, so ways = climbStairs(n-1) + climbStairs(n-2)
if(n==0) return 0;
else if(n == 1) return 1;
int [] ways = new int[n+1];
ways[1] = 1;
ways[2] = 2;
for(int i=3; i<ways.length; i++) {
ways[i] = ways[i-1] + ways[i-2];
}
return ways[n];
}
}