70. Climbing Stairs

Leetcode Dynamic Programming

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. 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

分析

这道题目考查的是基本的动态规划思想。设想你在楼梯的前$i$个台阶有numOfWays[i]种走法，那么前$i+1$个台阶有几种走法。分两种情况，可能此时你站在第$i$个台阶，那么向前一步即可；也有可能你站在第$i-1$个台阶，那么向前走两步即可。如果提前知道站在第$i$个台阶的走法数，和站在第$i-1$个台阶的走法数，那么把它们相加不就是前$i+1$个台阶的走法数。也就是

numOfWays[i] = numOfWays[i - 1] + numOfWays[i - 2];

根据上面的关系式，代码就非常简单了。

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

可以不存储中间结果：

public int climbStairs(int n) {
    if (n == 1) return 1;
    int last = 1, secondLast = 1, now = 0;
    for (int i = 2; i < n + 1; i++) {
        now = last + secondLast;
        secondLast = last;
        last = now;
    }
    return now;
}