<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[进击的ITgo]]></title><description><![CDATA[一只专研JavaEE和大数据技术的后端程序猿]]></description><link>https://blog.youkuaiyun.com/JSUTDOIT</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; JSUTDOIT]]></copyright><item><title><![CDATA[个人作品]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104363778</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104363778</guid><author>JSUTDOIT</author><pubDate>Mon, 17 Feb 2020 20:04:51 +0800</pubDate><description><![CDATA[Z部落:

  作品地址: https://dwz.cn/nGzR4iEm

  作品简介: 针对学校校区分散，各学院间交流不便的情况开发的一款校园新鲜事头条产品。项目得到校团委的大力支持。项目支持身份认证、邮件注册、新鲜事分享以及热门头条、话题栏目系统、新鲜事回复、点赞、收藏、楼中楼回复、新鲜事搜索、粉丝功能、消息推送、即时通讯、用户timeline以及详细主页列表、商家广告推荐...]]></description><category></category></item><item><title><![CDATA[序列化二叉树]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104242487</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104242487</guid><author>JSUTDOIT</author><pubDate>Sun, 09 Feb 2020 23:26:57 +0800</pubDate><description><![CDATA[题目链接:https://www.nowcoder.com/practice/cf7e25aa97c04cc1a68c8f040e71fb84?tpId=13&amp;tqId=11214&amp;tPage=1&amp;rp=1&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking




pub...]]></description><category></category></item><item><title><![CDATA[二叉搜索树的第k个结点]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104242047</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104242047</guid><author>JSUTDOIT</author><pubDate>Sun, 09 Feb 2020 22:43:57 +0800</pubDate><description><![CDATA[题目连接:https://www.nowcoder.com/practice/ef068f602dde4d28aab2b210e859150a?tpId=13&amp;tqId=11215&amp;tPage=4&amp;rp=4&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking




imp...]]></description><category></category></item><item><title><![CDATA[二叉树的深度]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104241922</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104241922</guid><author>JSUTDOIT</author><pubDate>Sun, 09 Feb 2020 22:33:50 +0800</pubDate><description><![CDATA[题目链接:https://www.nowcoder.com/practice/435fb86331474282a3499955f0a41e8b?tpId=13&amp;tqId=11191&amp;tPage=1&amp;rp=1&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking




imp...]]></description><category></category></item><item><title><![CDATA[二叉搜索树与双向链表]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104241555</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104241555</guid><author>JSUTDOIT</author><pubDate>Sun, 09 Feb 2020 22:28:58 +0800</pubDate><description><![CDATA[题目链接:

https://www.nowcoder.com/practice/947f6eb80d944a84850b0538bf0ec3a5?tpId=13&amp;tqId=11179&amp;tPage=1&amp;rp=1&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking

解题思路...]]></description><category></category></item><item><title><![CDATA[二叉搜索树的后序遍历序列]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104240622</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104240622</guid><author>JSUTDOIT</author><pubDate>Sun, 09 Feb 2020 21:03:46 +0800</pubDate><description><![CDATA[题目连接:https://www.nowcoder.com/practice/a861533d45854474ac791d90e447bafd?tpId=13&amp;tqId=11176&amp;tPage=1&amp;rp=1&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking

根据二叉搜索...]]></description><category></category></item><item><title><![CDATA[二叉树的镜像]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104163079</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104163079</guid><author>JSUTDOIT</author><pubDate>Mon, 03 Feb 2020 22:31:03 +0800</pubDate><description><![CDATA[题目链接:

递归交换左右部分


public class Solution {
    public void Mirror(TreeNode root) {
        if(root == null){
            return;
        }
        TreeNode tmp;
        tmp = root.left;
        root.l...]]></description><category></category></item><item><title><![CDATA[从上往下打印二叉树]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104163059</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104163059</guid><author>JSUTDOIT</author><pubDate>Mon, 03 Feb 2020 22:29:42 +0800</pubDate><description><![CDATA[题目链接:https://www.nowcoder.com/practice/7fe2212963db4790b57431d9ed259701?tpId=13&amp;tqId=11175&amp;tPage=1&amp;rp=1&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking



二叉树的...]]></description><category></category></item><item><title><![CDATA[树的子结构]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104162725</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104162725</guid><author>JSUTDOIT</author><pubDate>Mon, 03 Feb 2020 22:05:15 +0800</pubDate><description><![CDATA[题目链接:https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&amp;tqId=11170&amp;tPage=1&amp;rp=1&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking




/**...]]></description><category></category></item><item><title><![CDATA[重建二叉树]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104162273</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104162273</guid><author>JSUTDOIT</author><pubDate>Mon, 03 Feb 2020 21:28:14 +0800</pubDate><description><![CDATA[题目链接:https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=13&amp;tqId=11157&amp;tPage=1&amp;rp=1&amp;ru=/ta/coding-interviews&amp;qru=/ta/coding-interviews/question-ranking


import...]]></description><category></category></item><item><title><![CDATA[不同路径]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104129270</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104129270</guid><author>JSUTDOIT</author><pubDate>Fri, 31 Jan 2020 22:54:18 +0800</pubDate><description><![CDATA[题目链接:https://leetcode-cn.com/problems/unique-paths/




class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp =  new int[m][n];
         
        for(int i = 0;i &lt; m; i++)...]]></description><category></category></item><item><title><![CDATA[最小路径和]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104129083</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104129083</guid><author>JSUTDOIT</author><pubDate>Fri, 31 Jan 2020 22:34:45 +0800</pubDate><description><![CDATA[题目链接:

https://leetcode-cn.com/problems/minimum-path-sum/submissions/


class Solution {
    public int minPathSum(int[][] grid) {
        
         int m = grid.length;
         int n = grid[0].lengt...]]></description><category></category></item><item><title><![CDATA[打家劫舍 II]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104128683</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104128683</guid><author>JSUTDOIT</author><pubDate>Fri, 31 Jan 2020 21:56:32 +0800</pubDate><description><![CDATA[题目链接:https://leetcode-cn.com/problems/house-robber-ii/

本题与打家劫舍I相比 多了条件就是 环形相连，意味着第一个和最后一个 ，也只能选择其中之一。

 


class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0 || nums == ...]]></description><category></category></item><item><title><![CDATA[打家劫舍]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104127959</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104127959</guid><author>JSUTDOIT</author><pubDate>Fri, 31 Jan 2020 20:53:27 +0800</pubDate><description><![CDATA[题目链接:https://leetcode-cn.com/problems/house-robber/submissions/



思路:

  假设到第i家的时候可以偷得的最大值是:  dp[i] = Max{ dp[i-2] + nums[i] ,dp[i-1] }

  由于相邻的不能偷，所以结果为两者的最大值。


class Solution {
    public in...]]></description><category></category></item><item><title><![CDATA[存在重复元素]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104091030</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104091030</guid><author>JSUTDOIT</author><pubDate>Mon, 27 Jan 2020 09:54:19 +0800</pubDate><description><![CDATA[题目链接:https://leetcode-cn.com/problems/contains-duplicate/

利用哈希表破判重。


class Solution {
    public boolean containsDuplicate(int[] nums) {
        HashSet&lt;Integer&gt; set = new HashSet&lt;&gt;();
 ...]]></description><category></category></item><item><title><![CDATA[TinyURL 的加密与解密]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104091000</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104091000</guid><author>JSUTDOIT</author><pubDate>Mon, 27 Jan 2020 09:47:51 +0800</pubDate><description><![CDATA[题目链接:https://leetcode-cn.com/problems/encode-and-decode-tinyurl/description/

该题是hash表的一种应用。

随机固定长度加密 ，在这种方法中，使用数字和字母表集合来为 URL 生成加密结果。这种方法中，加密后的长度固定是 6 位。如果产生出来的加密结果与之前产生的结果一样，就换一个新的加密结果。


public cl...]]></description><category></category></item><item><title><![CDATA[两数之和]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104090995</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/104090995</guid><author>JSUTDOIT</author><pubDate>Mon, 27 Jan 2020 09:42:46 +0800</pubDate><description><![CDATA[题目链接:https://leetcode-cn.com/problems/two-sum/

该题是hash表的一种应用。


class Solution {
    public int[] twoSum(int[] nums, int target) {
        
        Map&lt;Integer,Integer&gt; map = new HashMap&lt;&gt...]]></description><category></category></item><item><title><![CDATA[环形链表1&环形链表2--双指针法的应用]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/103988677</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/103988677</guid><author>JSUTDOIT</author><pubDate>Wed, 15 Jan 2020 14:22:13 +0800</pubDate><description><![CDATA[环形链表1：

https://leetcode-cn.com/problems/linked-list-cycle/


方法一：哈希表

思路

我们可以通过检查一个结点此前是否被访问过来判断链表是否为环形链表。常用的方法是使用哈希表。

算法

我们遍历所有结点并在哈希表中存储每个结点的引用（或内存地址）。如果当前结点为空结点 null（即已检测到链表尾部的下一个结点），那么我们已经遍历...]]></description><category></category></item><item><title><![CDATA[删除链表的倒数第N个节点]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/103988169</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/103988169</guid><author>JSUTDOIT</author><pubDate>Wed, 15 Jan 2020 13:56:40 +0800</pubDate><description><![CDATA[方法一：两次遍历算法

思路

我们注意到这个问题可以容易地简化成另一个问题：删除从列表开头数起的第 (L - n + 1)个结点，其中 L是列表的长度。只要我们找到列表的长度 L，这个问题就很容易解决。

算法

首先我们将添加一个哑结点作为辅助，该结点位于列表头部。哑结点用来简化某些极端情况，例如列表中只含有一个结点，或需要删除列表的头部。在第一次遍历中，我们找出列表的长度 L。然后设置一个...]]></description><category></category></item><item><title><![CDATA[相交链表]]></title><link>https://blog.youkuaiyun.com/JSUTDOIT/article/details/103985687</link><guid>https://blog.youkuaiyun.com/JSUTDOIT/article/details/103985687</guid><author>JSUTDOIT</author><pubDate>Wed, 15 Jan 2020 11:20:52 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/intersection-of-two-linked-lists/


方法一: 暴力法

对链表A中的每一个结点ai，遍历整个链表 B 并检查链表 B 中是否存在结点和ai相同。

复杂度分析

时间复杂度 : (mn)。
空间复杂度 : O(1)O(1)。



方法二: 哈希表法

遍历链表 A 并将每个结点的地址/...]]></description><category></category></item></channel></rss>