70. Climbing Stairs

Description

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.

Solution

非常基础的动态规划题。

DP

dp[i]表示跳上第i级台阶有多少种方式。那么dp[i + 1] = dp[i] + dp[i - 1]。

public class Solution {
    public int climbStairs(int n) {
        int[] ways = new int[n + 1];
        ways[0] = 1;
        ways[1] = 1;
        
        for (int i = 2; i <= n; ++i) ways[i] = ways[i - 1] + ways[i - 2];
        
        return ways[n];
    }
}

DP with constant space

因为其实只需要记录prev以及prev.prev的值就可以了，用两个变量存储足矣。

public class Solution {
    public int climbStairs(int n) {
        int curr = 1, prev = 1;
        
        while (--n > 0) {
            int tmp = curr;
            curr += prev;
            prev = tmp;
        }
        
        return curr;
    }
}

更简化的版本，不需要tmp变量。

public class Solution {
    public int climbStairs(int n) {
        int curr = 1, prev = 1;
        while (--n > 0) prev = (curr += prev) - prev;
        return curr;
    }
}