[LeetCode] 70. Climbing Stairs

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Problem

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:

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Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

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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

Explanation

  1. We can use dynamic programming to solve this problem. The base case is when n = 0, res = 0; when n = 1, res = 1; when n = 2, res = 2.

  2. The way to reach the current i step is equal to the way to reach i-1 step plus the way to reach i-2 step.

Solution

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class Solution {
    public int climbStairs(int n) {
        if (n <= 2) return n;
        int pre = 1;
        int cur = 2;
        for (int i = 3; i <= n; i++) {
            cur = cur + pre;
            pre = cur - pre;
        }
        return cur;
    }
}

// class Solution {
//     public int climbStairs(int n) {
//         if (n <= 2) return n;
//         int[] dp = new int[n+1];
//         dp[0] = 0;
//         dp[1] = 1;
//         dp[2] = 2;
//         for (int i = 3; i <= n; i++) {
//             dp[i] = dp[i-1] + dp[i-2];
//         }
//         return dp[n];
//     }
// }