soulmachine / leetcode

LeetCode题解，151道题完整版。广告：推荐刷题网站 https://www.lintcode.com/?utm_source=soulmachine

LeetCode题解

在线阅读

https://algorithm-essentials.soulmachine.me/

PDF下载

LeetCode题解(C++版).pdf

C++ 文件夹下是C++版，内容一模一样，代码是用C++写的。

Java 文件夹下是Java版，目前正在编写中，由于拖延症，不知道猴年马月能完成。

如何编译PDF

命令行编译

docker run -it --rm -v $(pwd)/C++:/project -w /project soulmachine/texlive xelatex -interaction=nonstopmode leetcode-cpp.tex

vscode下编译

本项目已经配置好了vscode devcontainer, 可以在 Windows, Linux 和 macOS 三大平台上编译。

用 vscode 打开本项目，选择右下角弹出的 "Reopen in Container"，就会在容器中打开本项目，该容器安装了 Tex Live 2022 以及所需要的10个字体。

点击vscode左下角的齿轮图标，选择 Command Palette，输入tasks, 选择 Run Task， 选择 leetcode-C++，即可启动编译。

LaTeX模板

本书使用的是陈硕开源的模板。这个模板制作精良，感谢陈硕 :)

这个LaTex模板总共使用了10个字体，下载地址 https://pan.baidu.com/s/1eRFJXnW 。有的字体Windows自带了，有的字体Ubuntu自带了，但都不全，还是一次性安装完所有字体比较方便。

也可以参考 Dockerfile 去学习如何安装所有字体。

贡献代码

欢迎给本书添加内容或纠正错误，在自己本地编译成PDF，预览没问题后，就可以发pull request过来了。

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leetcode's Issues

关于Java版本

您好，Java目录下没有内容，是还没有来得做么？
我最近正在使用您总结的教材来刷题，我是用Java的，可以正好补充Java的版本。

The solution of Regular Expression Matching is not O(N)

The recursive version can be accepted on leetcode, but it's not a O(N) solution.
Please consider cases from wildcard matching:

abbabaaabbabbaababbabbbbbabbbabbbabaaaaababababbbabababaabbababaabbbbbbaaaabababbbaabbbbaabbbbababababbaabbaababaabbbababababbbbaaabbbbbabaaaabbababbbbaababaabbababbbbbababbbabaaaaaaaabbbbbaabaaababaaaabb
.*aa.*ba.*a.*bb.*aa.*ab.*a.*aaaaaa.*a.*aaaa.*bba

This case never return on my machine.

LeetCode题解c++版本的P44页有问题

P44页讲到了Remove Duplicates from Sorted List的迭代解法，其中的一段代码片段：

for (ListNode *prev = head, *cur = head->next; cur; cur = cur->next) {
    if (prev->val == cur->val) {
        prev->next = cur->next;
        delete cur;
    } else {
        prev = cur;
    }
}

既然delete cur语句已经将cur删除，为什么还可以在for的条件语句里执行cur = cur->next?

PS.感谢作者辛勤的劳动为我们带来了如此有价值的作品，再次感谢:）

Insertion Sort Link List C++

OJ没法通过, 如果linked list: 1->NULL or 1->1-> NULL

Regular Expression Matching

I do not think the time complexity of your code is O (n) but I am not sure what it should be.

Sudoku Solver 代码这里感觉有问题 (chapDFS, 900行)

bool isValid(const vector<vector<char> > &board, int x, int y) {
    int i, j;
    for (i = 0; i < 9; i++) // 检查 y 列
        if (i != x && board[i][y] == board[x][y])
            return false;
    for (j = 0; j < 9; j++) // 检查 x 行
        if (j != y && board[x][j] == board[x][y])
            return false;
    for (i = 3 * (x / 3); i < 3 * (x / 3 + 1); i++)
        for (j = 3 * (y / 3); j < 3 * (y / 3 + 1); j++)
            if (i != x && j != y && board[i][j] == board[x][y])
            -------------------------------------------------
             这个地方, 我感觉应该是 if ((i != x || j != y) &&  board[i][j] == board[x][y]) ?
             应该只是排除board[x][y] 自身以后, 然后看有没有和他重复的吧.

                return false;
    return true;
}

String to Integer don't take care of '+-1' case

Hi, Your solution don't take care this case: '+-1'.
I am working on python solutions. I will submit a pr for this after I finish my work(weeks later).
This is a reminder. Or you can take care of it yourself.

15.13 divide two integers, need to retrun INT_MAX when result is overflow

Leetcode has added the new case for this problem, when overflow occurs, need to return INT_MAX, the current code can NOT be accepted. I adjust the return code as this and accepted.

    long long ret =    ((dividend^divisor) >> 31)?(-multi): multi;

    if (ret > INT_MAX)
        return INT_MAX;
    else
        return ret;

发现一个错别字

第一章编程技巧里“阀值”应该为“阈值”，这个在知乎上看到过两者区别。
http://www.zhihu.com/question/20642950

c++版本3.4题（add binary）的空间复杂度应该是O(n)吧

surrounded regions 貌似有bug

for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)

应该改成
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)

至少我目前测试的结果是这样。

Symmetric Tree递归算法

_解题思路: 递归
这道题没什么特别的地方，现在这里简单的分析一下解题思路，从根节点往下，我们要判断三个条件.

1. 左右两个节点的大小是否相同.
2. 左节点的左孩子是否和右节点的右孩子相同.
3. 左节点的右孩子是否和右节点的左孩子相同.
   ，如果以上三个条件对于每一层都满足，我们就可以认为这棵树是镜像树._

但这并不是树是对称树的所有条件，比如
1
/
2 2
/ \ /
1 2 2 1
/ \ / \ / \ /
1 3 3 1 1 2 2 1

所有的层都符合这三个条件，但显然不是对称树
下一层满足了条件，并不等于上层也满足。

An optimized sol for 10.2.4

2ms with the following sol.
Optimized the combination part.

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(!m || !n) return 0;

        long long a = 1,b = 1;

        // find the max
        int min = m < n ? m : n;
        int max = m > n ? m : n;

        // find (m+n-2)!/max{(m-1)!,(n-1)!}
        for(int i=max;i<=m+n-2;++i)  a *= i;
        // min{(m-1)!,(n-1)!}
        for(int i=1;i<=min-1;++i)    b *= i;

        return a/b;
    }
};

longest palindromic substring

https://github.com/soulmachine/leetcode/blob/master/C%2B%2B/chapString.tex#L402
the iteration direction for inner loop is not correct and should be
for (int j = i - 1; j >= 0 ; j--) {

Symmetric Tree

The problems description is the same as Same Tree
=.=

是“阈值”不是“阀值”

第一章第一段：“应判断两者之差的绝对值f(a,b)是否小于某个阀值“。其中“阀值”是一个错误的用词，正确的用法是“阈值”。“阈“音”玉“，其内部是”或“不是”伐“。

标题上的链接“该页无法显示”

http://q.weibo.com/1312378

字体链接失效

字体的百度网盘链接已经失效了，能不能重新补一个？

A problem about "3 Sum"

About insert interval

Time limit exceeded.

linear solution:
https://oj.leetcode.com/discuss/20176/concise-in-place-solution-with-linear-time

Binary search:
https://oj.leetcode.com/discuss/19000/c-my-binary-search-solution

2.1.2 Remove Duplicates from Sorted Array II

Code 2: There isn't any occur variable in this solution. I am confused by what you are talking about.

15.3 Insert Interval Time Limit Exceeded

Hi, I gotta to run your solution on github.
However, I got a TLE when submit to online judge.
The corresponding testcase is something like
[[1,2],[3,4],[5,6],[7,8],[9,10],[11,12],[13,14],[15,16],[17,18] ... [19999,20000]], [0,20001]]
It cause multiple erase which make your algorithm in worst case O(n^2).
A O(n) time, O(n) space solution that use extra memory to save the result can basically pass the oj.

请问懒人镜像版的root密码是什么啊？

请问懒人镜像版的root密码是什么啊？

15.1 Reverse Integer(cpp): have not check integer overflow....

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?
yeah?

Error in link address

in 2.1.18, gray code, the link to the wikipedia page is wrong.

the link points to http://en.wikipedia.org/wiki/Gray_coded which should be http://en.wikipedia.org/wiki/Gray_code

Problem in p35 - Gas station

Line 16: return total>=0 ? j+1 : -1
Should be (j+1)%gas.size(), so that we will return 0 instead of N in some cases.

Flatten Binary Tree to Linked List 要求"in place"

这个题目要求in-place，但目前的解好像都不是空间O(1)的。我有个O(1)空间的解想放上来但看到其他帖子里说这个repo是private了。现在可以pull request了么？

15.1 为啥要用long long?

Longest Substring Without Repeating Characters

The test set also includes the non-alphabet characters, thus the ASCII_MAX should be defined as 256, and s[i] - 'a' is not necessary.
The current code can not pass the OJ. I made minor changes as below with accepted:

class Solution {
public:
int lengthOfLongestSubstring(string s) {
const int ASCII_MAX= 256;
int last_pos[ASCII_MAX];
fill(last_pos, last_pos+ASCII_MAX, -1);
int max_len = 0, len = 0;
for(int i=0; i<s.size(); i++,len++){
if(last_pos[s[i]] >= 0){//find a duplicated char
max_len = max(len, max_len);
i = last_pos[s[i]] + 1;//start from the next possible solution
len = 0;
fill(last_pos, last_pos+ASCII_MAX, -1);
}
last_pos[s[i]] = i;
}
max_len = max(len, max_len);
return max_len;
}
};

Difference between p == nullptr and !p

Suppose p is a pointer. The running time for if(p==nullptr) and if(!p) are different in Leetcode. Is this a bug in Leetcode.

注释错误

问题word break2 里
// path[i][j] 为 true,表示 s[j, i) 是一个合法单词,可以从 j 处切开
// 第一行未用
vector<vector > prev(s.length() + 1, vector(s.length()));

这里的path[i][j]为true....应该改为prev[i][j]...

The time complexity of Valid Sudoku may be wrong.

it says O(N^2), but I think it should be O(1) since the size is fixed.

5.1.9 Symmetric Tree这个题目描述是不是错了

原题描述是
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:
1
/
2 2
/ \ /
3 4 4 3

But the following is not:
1
/
2 2
\
3 3

仓主现在是不打算继续更新了吗？

觉得很可惜，这个题解很受用，希望可以继续更下去

Scramble string complexity

I have a doubt about the complexity of the recursive method. This should not be n^6. The worst case should be 6^n instead of n^6.

Merge k Sorted Lists cannot pass

Using function mergeTwoLists will lead to Time limit exceeding. Please use min heap.

English version of book

12.5 Longest Substring Without Repeating Characters

The input string could contain characters other than a-z. So we need to change the size of last array from 26 to 256.

Recursive solution for 2.2.8 Swap Nodes in Pairs

ListNode* swapPairs(ListNode *head) {
  if(head == NULL || head->next == NULL)
    return head;

  ListNode *next = head->next;
  ListNode *temp = next->next;
  next->next = head;
  head->next = swapPairs(temp);

  return next;
}

NumTrees 一题的解法似乎有改进余地

N 个节点的 BST 的形态为 Catalan 数的第 N 项，即 C(2n, n) / (n + 1) => (2n)! / n! / (n + 1)，最后可以写成这样：

class Solution {
 public:
  int numTrees(int n) {
    long ans = 1;
    for (int i = 1; i <= n; ++i) ans += ans * n / i;
    return ans / (n + 1);
  }
};

Merge k Sorted Lists

解决了超时的问题。应该使用multi-pass来两两merge

leetcode c++

这里包含一些新题，需要补充

leetcode在陆续增加新题。持续更新：https://github.com/ghostrong/leetcode, C++源码, .cpp 文件

仓主不理这个仓库了？

RT

there is a problem about "Two Sum"

In a case the code is not right: if gap == num[i] then you will get result[0]=result[1]

LRU Cache: Not necessary to update cacheMap after splice

In LRU Cache, after moving a node in the list to the front, it updated cacheMap[key] to cacheList.begin().

But I don't think the iterator to the node will be changed after splicing.

A doubt on p36 - candy

++inc may be better, although inc++ could AC.

Validate Binary Search Tree

Consider the case: Only root node, whose value is INT_MAX. The code will output false instead of true.

Solution: Use LLONG_MAX to replace INT_MAX.

class Solution {
public:
bool isValidBST(TreeNode* root) {
return isValidBST(root, LLONG_MIN, LLONG_MAX);
}
bool isValidBST(TreeNode* root, long long lower, long long upper) {
if (!root) return true;
return root->val > lower && root->val < upper
&& isValidBST(root->left, lower, root->val)
&& isValidBST(root->right, root->val, upper);
}
};

leetcode/C++/chapDFS.tex line263 seems typo?

In the end of

leetcode/C++/chapDFS.tex line263
一个$m$行，$n$列的矩阵，机器人从左上走到右下总共需要的步数是$m+n-2$，其中向下走的步数是$m-1$，因此问题变成了在$m+n-2$个操作中，选择$m–1$个时间点向下走，选择方式有多少种。即 $C_{m+n-2}^{m-1}$ 。

$C_{m+n-2}^{m-1} would present like

m - 1
C m + n - 2

But it should be

m + n - 2
C m - 1

Right? Or your country have different usage of C(m, n)?

If it is a typo, please change it to

一个$m$行，$n$列的矩阵，机器人从左上走到右下总共需要的步数是$m+n-2$，其中向下走的步数是$m-1$，因此问题变成了在$m+n-2$个操作中，选择$m–1$个时间点向下走，选择方式有多少种。即 $C_{m-1}^{m+n-2}$

Leetcode Java Sqrt(x) 题代码错误

提供的代码似乎是题目Plus One的答案，和题目Sqrt(x)没有关系