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Blankj/awesome-java-leetcode
树的打印函数好像对null处理有点问题，麻烦大佬帮看下

例如Leetcode94题，输入: [1,null,2,3]，调用TreeNode.print( )函数，3会丢失
  • 1580923067
  • 1
  • Opened 
    on Aug 8, 2020
  • #36

Blankj/awesome-java-leetcode
招聘贴：阿里巴巴消息业务/基础团队招聘（校招/社招）

我们团队都是Java哈，也有cpp、ios和android，大家刷好题了就可以随时来碾压我们了。欢迎大家加入。 招聘, 微信：kerisle， 大家也可以直接发邮件 hepin.p@alibaba-inc.com image
  • He-Pin
  • Opened 
    on Mar 9, 2020
  • #33

Blankj/awesome-java-leetcode
看了楼主的镇楼诗，我果断的点了个star

  • mxdldev
  • 1
  • Opened 
    on Sep 5, 2019
  • #28

Blankj/awesome-java-leetcode
two num

两个数相加，如果是Integer的情况，用一个方法处理该如何操作呢？
  • Mrmarktoo
  • 1
  • Opened 
    on May 24, 2019
  • #27

Blankj/awesome-java-leetcode
Longest Common Prefix

如果有N个字符串，那就需要N个循环了啊
  • ranxiaoyi
  • 1
  • Opened 
    on Apr 11, 2019
  • #25

Blankj/awesome-java-leetcode
003

private static int lengthOfLongestSubstringMe(String s) { HashMap Character, Integer map = new HashMap (128); int max = 0; int pre = 0; char temp; for (int i = 0; i s.length(); i++) { temp = s.charAt(i); ...
  • MohnSnow
  • Opened 
    on Dec 3, 2018
  • #24

Blankj/awesome-java-leetcode
/note/043/README.md Multiply Strings

class Solution { public String multiply(String num1, String num2) { if (num1.equals( 0 ) || num2.equals( 0 )) return 0 ; int l1 = num1.length(), l2 = num2.length(), l = l1 + l2; ...
  • imckh
  • 1
  • Opened 
    on Nov 24, 2018
  • #23

Blankj/awesome-java-leetcode
67.二进制求和，重复判断

不需要在循环里再判断p1,p2是否大于等0了，循环条件就是p1 =0 p2 =0。 while (p1 = 0 p2 = 0) { //carry += p1 = 0 ? a.charAt(p1--) - 0 : 0; //carry += p2 = 0 ? b.charAt(p2--) - 0 : 0; carry += a.charAt(p1--) - 0 ; carry += b.charAt(p2--) ...
  • xiaxinandye
  • 1
  • Opened 
    on Oct 23, 2018
  • #21

Blankj/awesome-java-leetcode
21、合并两个链表

LZ你好，请问一下为什么最后返回的是head.next而不是temp.next？
  • 24Kshign
  • 1
  • Opened 
    on Oct 12, 2018
  • #20

Blankj/awesome-java-leetcode
链表中的数值两两交换

class Solution { public ListNode swapPairs(ListNode head) { if (head == null || head.next == null) return head; ListNode node = head.next; head.next = swapPairs(node.next); node.next = head; return node; ...
  • kaikai-sk
  • 2
  • Opened 
    on Oct 11, 2018
  • #19
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