leetcode day01

最新推荐文章于 2025-02-10 16:59:43 发布

pigofzhou

最新推荐文章于 2025-02-10 16:59:43 发布

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分类专栏：算法编程题 leetcode

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本文精选LeetCode上的经典算法题目，包括求二叉树的最大深度、计算股票最大收益、查找只出现一次的数字及反转整数等。通过具体示例和代码实现，深入浅出地解析算法原理及技巧。

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时间限制：1秒空间限制：32768K 热度指数：13273

本题知识点：树 leetcode

算法知识视频讲解

题目描述

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode *root) {
        if(!root) return 0;
        int cnt=0;
        cnt=max(maxDepth(root->left),cnt);
        cnt=max(maxDepth(root->right),cnt);
        return cnt+1;
    }
};

时间限制：1秒空间限制：32768K 热度指数：5014

本题知识点：数组 leetcode

算法知识视频讲解

题目描述

Say you have an array for which the i th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int len=prices.size();
        if(len<=1) return 0;
        int res=0;
        for(int i=1;i<len;i++)
        {
            if(prices[i]-prices[i-1]>0)
                res+=prices[i]-prices[i-1];
        }
        return res;
    }
};

时间限制：1秒空间限制：32768K 热度指数：12968

本题知识点：复杂度 leetcode

算法知识视频讲解

题目描述

Given an array of integers, every element appears twiceexcept for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

class Solution {
public:
    int singleNumber(int A[], int n) {
        int res=0;
        for(int i=0;i<n;i++)
            res^=A[i];
        return res;
    }
};

时间限制：1秒空间限制：32768K 热度指数：4390

本题知识点：复杂度 leetcode

算法知识视频讲解

题目描述

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

click to show spoilers.

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

class Solution {
public:
    int reverse(int x) {
        int num=abs(x);
        int res=0;
        while(num)
        {
            res=res*10+num%10;
            num/=10;
        }
        if(x>0) return res;
        else return res*(-1);
    }
};