계단 수 세기 문제 (recursive + dynamic programming)

gistfile1.txt

//https://oj.leetcode.com/problems/climbing-stairs/

Q. 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?

//n=2는 n=1에서 1걸음 더 간것
//n=3은 n=2에서 1걸음 더 간것 + n=1에서 2걸음 
//n=4는 n=3에서 1걸음 더 간것 + n=2에서 2걸음 
// .. => 재귀함수로 n을 감소시키면서 풀면 되겠다.
// n>2, f(n) = f(n-1) + f(n-2) 성립
//f(3) = f(2) + f(1) = 3
//f(4) = f(3) + f(2)  = 3 + 2 = 5
//time over : dynamic programming 적용하자.

class Solution {
public:
    int steps[1024];
    int climbStairs(int n) {
        if(n == 1) {
            if(steps[1] == 0) {
                steps[1] = 1;
                return steps[1];
            } else {
                return steps[1];
            }
        }
        else if(n == 2) {
            if(steps[2] == 0) {
                steps[2] = 2;
            return steps[2];
            } else {
                return steps[2];
            }
        }
        else if(n >= 3) {
            if(steps[n] == 0) {
                steps[n] = climbStairs(n-1) + climbStairs(n-2);
                return steps[n];
            } else {
                return steps[n];   
            }
        }
    }
};


//복습 
int steps[1024];
int countway(int n) {
    if(n <= 1) {
        steps[n] = 1;
        return steps[n];
    }
    
    if(n == 2) {
        steps[n] = 2;
        return steps[n];
    }
    
    if(n >= 3) {
        if( steps[n] == 0 ) {
            steps[n] = countway(n-1) + countway(n-2);
        } 
        
        return steps[n]; 
    }
}