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70. Climbing Stairs.md

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70. Climbing Stairs

leetcode Daily Challenge on July 31th, 2020.

Difficulty : Easy

Related Topics : 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

Solution

same as 518. Coin Change 2.

  • mine
    • Java

      • Brute Force Time Limit Exceeded

        int res = 0;
public int climbStairs(int n) {
    next(0,n);
    return res;
}

public void next(int i, int sum){
    if(i < sum){
        next(i + 1, sum);
        next(i + 2, sum);
    }else if(i == sum){
        res++;
    }
}

      • Dynamic Programming Runtime: 0 ms, faster than 100.00%, Memory Usage: 35.8 MB, less than 5.26% of Java online submissions

        //O(N)time O(N)space
public int climbStairs(int n) {
    int[] res = new int[n + 1];
    res[0] = 1;
    res[1] = 1;
    for(int i = 2; i <= n; i++){
        res[i] += res[i-1] + res[i-2];
    }
    return res[n];
}
  • the leetcode solution
  • Recursion with Memoization Runtime: 0 ms, faster than 100.00%, Memory Usage: 35.8 MB, less than 5.26% of Java online submissions 
    //O(N)time O(N)space
public int climbStairs(int n) {
    int memo[] = new int[n + 1];
    return climb_Stairs(0, n, memo);
}
public int climb_Stairs(int i, int n, int memo[]) {
    if (i > n) {
        return 0;
    }
    if (i == n) {
        return 1;
    }
    if (memo[i] > 0) {
        return memo[i];
    }
    memo[i] = climb_Stairs(i + 1, n, memo) + climb_Stairs(i + 2, n, memo);
    return memo[i];
}