算法题之Climbing Stairs（leetcode 70）

题目：

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.

Approach #1 Brute Force [Time Limit Exceeded]

public class Solution {
    public int climbStairs(int n) {
        climb_Stairs(0, n);
    }
    public int climb_Stairs(int i, int n) {
        if (i > n) {
            return 0;
        }
        if (i == n) {
            return 1;
        }
        return climb_Stairs(i + 1, n) + climb_Stairs(i + 2, n);
    }
}

Time complexity : O(2^n). Size of recursion tree will be 2^n​​.

Space complexity : O(n). The depth of the recursion tree can go upto n.

 

Approach #2 Recursion with memorization [Accepted]

public class Solution {
    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];
    }
}

Time complexity : O(n). Size of recursion tree can go upto n.

Space complexity : O(n). The depth of recursion tree can go upto n.

 

Approach #3 Dynamic Programming [Accepted]

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

Time complexity : O(n). Single loop upto n.

Space complexity : O(n). dp array of size n is used.

 

Approach #4 Fibonacci Number [Accepted]:

public class Solution {
    public int climbStairs(int n) {
        if (n == 1) {
            return 1;
        }
        int first = 1;
        int second = 2;
        for (int i = 3; i <= n; i++) {
            int third = first + second;
            first = second;
            second = third;
        }
        return second;
    }
}

Time complexity : O(n). Single loop upto n is required to calculate n^{th} fibonacci number.

Space complexity : O(1). Constant space is used.

 

原文：https://leetcode.com/articles/climbing-stairs/