Leetcode70

Climbing Stairs:
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?
Note: Given n will be a positive integer.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1 step + 1 step
2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1 step + 1 step + 1 step
1 step + 2 steps
2 steps + 1 step

Solution_01: 递归方法

#include <iostream>

using namespace std;

// 递归的方法
class Solution {
private:
    int calcWays(int n){

        if ( n == 1)
            return 1;
        if ( n == 2)
            return 2;

        return calcWays(n-1) + calcWays(n-2);
    }
public:
    int climbStairs(int n) {

        return calcWays(n);

    }
};

Solution_02: 记忆化搜索

#include <iostream>
#include <vector>

using namespace std;

class Solution {
private:
    vector<int> memo;

    int calcWays(int n){

        if ( n == 1)
            return 1;
        if ( n == 2)
            return 2;

        if ( memo[n] == -1 )
             memo[n] = calcWays(n-1) + calcWays(n-2);

        return memo[n];
    }
public:
    int climbStairs(int n) {


        memo = vector<int>(n+1, -1);
        return calcWays(n);

    }
};

Solution_03: 动态规划

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    int climbStairs(int n) {

        vector<int> memo(n+1, -1);
        memo[1] = 1;
        memo[2] = 2;
        for ( int i = 3; i <= n; i++)
            memo[i] = memo[i-1] + memo[i-2];

        return memo[n];

    }
};

总结： 仔细分析这个问题，可以转换为斐波那契数列问题。