From 7011c075afba13e927c171924018f6600b1c7203 Mon Sep 17 00:00:00 2001 From: youngyangyang04 <826123027@qq.com> Date: Wed, 19 Aug 2020 10:33:07 +0800 Subject: [PATCH] Update --- README.md | 17 +++-- pics/93_复原IP地址.png | Bin 0 -> 44492 bytes ...83.删除排序链表中的重复元素.md | 2 +- problems/0093.复原IP地址.md | 63 ++++++++++++++++++ problems/0131.分割回文串.md | 49 ++++++++++++++ problems/0349.两个数组的交集.md | 31 ++++++--- problems/0383.赎金信.md | 9 +-- 7 files changed, 154 insertions(+), 17 deletions(-) create mode 100644 pics/93_复原IP地址.png create mode 100644 problems/0093.复原IP地址.md create mode 100644 problems/0131.分割回文串.md diff --git a/README.md b/README.md index be72f97b..9a5ad2cb 100644 --- a/README.md +++ b/README.md @@ -28,12 +28,18 @@ * [这六道哈希表相关的面试题，你一定要会！](https://mp.weixin.qq.com/s/nxuWv5cUhCPSbAdIHtWgSg) * [关于链表，你该了解这些！](https://mp.weixin.qq.com/s/ntlZbEdKgnFQKZkSUAOSpQ) * [刷leetcode的时候，究竟什么时候可以使用库函数，什么时候不要使用库函数，过来人来说一说](https://leetcode-cn.com/circle/article/E1Kjzn/) +* [链表：听说用虚拟头节点会方便很多？](https://mp.weixin.qq.com/s/slM1CH5Ew9XzK93YOQYSjA) +* [链表：一道题目考察了常见的五个操作！](https://mp.weixin.qq.com/s/Cf95Lc6brKL4g2j8YyF3Mg) +* [链表：听说过两天反转链表又写不出来了？](https://mp.weixin.qq.com/s/pnvVP-0ZM7epB8y3w_Njwg) +* [链表：环找到了，那入口呢？](https://mp.weixin.qq.com/s/_QVP3IkRZWx9zIpQRgajzA) +* [哈希表：可以拿数组当哈希表来用，但哈希值不要太大](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig) * 精选链表相关的面试题 * 精选字符串相关的面试题 * 精选栈与队列相关的面试题 * 精选二叉树相关的面试题 * 精选递归与回溯面试题 + （持续更新中....） # LeetCode 刷题攻略 @@ -50,10 +56,10 @@ * [0059.螺旋矩阵II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0059.螺旋矩阵II.md) * 链表经典题目 - * [0203.移除链表元素](https://github.com/youngyangyang04/leetcode/blob/master/problems/0203.移除链表元素.md) - * [0707.设计链表](https://github.com/youngyangyang04/leetcode/blob/master/problems/0707.设计链表.md) - * [0206.翻转链表](https://github.com/youngyangyang04/leetcode/blob/master/problems/0206.翻转链表.md) - * [0142.环形链表II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0142.环形链表II.md) + * [0203.移除链表元素](https://mp.weixin.qq.com/s/slM1CH5Ew9XzK93YOQYSjA) + * [0707.设计链表](https://mp.weixin.qq.com/s/Cf95Lc6brKL4g2j8YyF3Mg) + * [0206.翻转链表](https://mp.weixin.qq.com/s/pnvVP-0ZM7epB8y3w_Njwg) + * [0142.环形链表II](https://mp.weixin.qq.com/s/_QVP3IkRZWx9zIpQRgajzA) * 哈希表经典题目 * [0242.有效的字母异位词](https://github.com/youngyangyang04/leetcode/blob/master/problems/0242.有效的字母异位词.md) @@ -335,6 +341,7 @@ int countNodes(TreeNode* root) { |---|---| ---| --- | |[0001.两数之和](https://github.com/youngyangyang04/leetcode/blob/master/problems/0001.两数之和.md) | 数组|简单|**暴力** **哈希**| |[0015.三数之和](https://github.com/youngyangyang04/leetcode/blob/master/problems/0015.三数之和.md) | 数组 |中等|**双指针** **哈希**| +|[0017.电话号码的字母组合](https://github.com/youngyangyang04/leetcode/blob/master/problems/0017.电话号码的字母组合.md) | 回溯 |中等|**回溯**| |[0018.四数之和](https://github.com/youngyangyang04/leetcode/blob/master/problems/0018.四数之和.md) | 数组 |中等|**双指针**| |[0020.有效的括号](https://github.com/youngyangyang04/leetcode/blob/master/problems/0020.有效的括号.md) | 栈 |简单|**栈**| |[0021.合并两个有序链表](https://github.com/youngyangyang04/leetcode/blob/master/problems/0021.合并两个有序链表.md) |链表 |简单|**模拟** | @@ -345,6 +352,7 @@ int countNodes(TreeNode* root) { |[0053.最大子序和](https://github.com/youngyangyang04/leetcode/blob/master/problems/0053.最大子序和.md) |数组 |简单|**暴力** **贪心** 动态规划 分治| |[0059.螺旋矩阵II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0059.螺旋矩阵II.md) |数组 |中等|**模拟**| |[0083.删除排序链表中的重复元素](https://github.com/youngyangyang04/leetcode/blob/master/problems/0083.删除排序链表中的重复元素.md) |链表 |简单|**模拟**| +|[0093.复原IP地址](https://github.com/youngyangyang04/leetcode/blob/master/problems/0093.复原IP地址) |回溯 |中等|**回溯**| |[0094.二叉树的中序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0094.二叉树的中序遍历.md) |树 |中等|**递归** **迭代/栈**| |[0098.验证二叉搜索树](https://github.com/youngyangyang04/leetcode/blob/master/problems/0098.验证二叉搜索树.md) |树 |中等|**递归**| |[0100.相同的树](https://github.com/youngyangyang04/leetcode/blob/master/problems/0100.相同的树.md) |树 |简单|**递归** | @@ -353,6 +361,7 @@ int countNodes(TreeNode* root) { |[0104.二叉树的最大深度](https://github.com/youngyangyang04/leetcode/blob/master/problems/0104.二叉树的最大深度.md) |树 |简单|**递归** **迭代/队列/BFS**| |[0110.平衡二叉树](https://github.com/youngyangyang04/leetcode/blob/master/problems/0110.平衡二叉树.md) |树 |简单|**递归**| |[0111.二叉树的最小深度](https://github.com/youngyangyang04/leetcode/blob/master/problems/0111.二叉树的最小深度.md) |树 |简单|**递归** **队列/BFS**| +|[0131.分割回文串](https://github.com/youngyangyang04/leetcode/blob/master/problems/0131.分割回文串.md) |回溯 |中等|**回溯**| |[0142.环形链表II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0142.环形链表II.md) |链表 |中等|**快慢指针/双指针**| |[0144.二叉树的前序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0144.二叉树的前序遍历.md) |树 |中等|**递归** **迭代/栈**| |[0145.二叉树的后序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0145.二叉树的后序遍历.md) |树 |困难|**递归** 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更多算法干货文章持续更新，可以微信搜索「代码随想录」第一时间围观，关注后，回复「Java」「C++」 「python」「简历模板」「数据结构与算法」等等，就可以获得我多年整理的学习资料。 diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md new file mode 100644 index 00000000..66ad1ec6 --- /dev/null +++ b/problems/0093.复原IP地址.md @@ -0,0 +1,63 @@ +## 题目地址 +https://leetcode-cn.com/problems/restore-ip-addresses/ + +## 思路 + + +## C++代码 + +``` +class Solution { +private: + vector result;// 记录结果 + // startIndex: 搜索的起始位置，pointNum:添加逗点的数量 + void search(string& s, int startIndex, int pointNum) { + if (pointNum == 3) { // 逗点数量为3时，分隔结束 + // 判断第四段子字符串是否合法，如果合法就放进result中 + if (isValid(s, startIndex, s.size() - 1)) { + result.push_back(s); + } + return; + } + // 从起始位置开始构造字段字符串串 + for (int i = startIndex; i < s.size(); i++) { + // 判断 [startIndex,i] 这个区间的子串是否合法 + if (isValid(s, startIndex, i)) { + // 合法，在i的后面插入一个逗点 + s.insert(s.begin() + i + 1 , '.'); + // 插入逗点之后下一个子串的起始位置为i+2 + search(s, i + 2, pointNum + 1); + s.erase(s.begin() + i + 1); // 回溯时删掉逗点 + } else break; + } + } + // 判断字符串s在左闭又闭区间[start, end]所组成的数字是否合法 + bool isValid(const string& s, int start, int end) { + if (start > end) { + return false; + } + if (s[start] == '0' && start != end) {// 0开头的数字不合法 + return false; + } + int num = 0; + for (int i = start; i <= end; i++) { + if (s[i] > '9' || s[i] < '0') { // 遇到非数字字符不合法 + return false; + } + num = num * 10 + (s[i] - '0'); + if (num > 255) { // 如果大于255了不合法 + return false; + } + } + return true; + } +public: + vector restoreIpAddresses(string s) { + result.clear(); + search(s, 0, 0); + return result; + } +}; +``` + +> 更多算法干货文章持续更新，可以微信搜索「代码随想录」第一时间围观，关注后，回复「Java」「C++」 「python」「简历模板」「数据结构与算法」等等，就可以获得我多年整理的学习资料。 diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md new file mode 100644 index 00000000..9cae7bdd --- /dev/null +++ b/problems/0131.分割回文串.md @@ -0,0 +1,49 @@ + +## 题目地址 +https://leetcode-cn.com/problems/palindrome-partitioning/ + +## 思路 + + +## C++代码 +``` +class Solution { +private: + vector> result; + + // startIndex: 搜索的起始位置，vec: 放已经回文的子串 + void search(const string& s, int startIndex, vector& vec) { + // 如果起始位置已经大于s的大小，说明已经找到了一组分割方案了 + if (startIndex >= s.size()) { + result.push_back(vec); + return; + } + for (int i = startIndex; i < s.size(); i++) { + if (isPalindrome(s, startIndex, i)) { // 是回文子串 + // 获取[startIndex,i]在s中的子串 + string str = s.substr(startIndex, i - startIndex + 1); + vec.push_back(str); + } else { // 如果不是则直接跳过 + continue; + } + search(s, i + 1, vec); // 寻找i+1为起始位置的子串 + vec.pop_back(); // 回溯过程，弹出本次已经填在的子串 + } + } + bool isPalindrome(const string& s, int start, int end) { + for (int i = start, j = end; i < j; i++, j--) { + if (s[i] != s[j]) { + return false; + } + } + return true; + } +public: + vector> partition(string s) { + result.clear(); + vector vec; + search(s, 0 , vec); + return result; + } +}; +``` diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 2f36c503..408ef931 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -2,21 +2,35 @@ ## 题目地址 https://leetcode-cn.com/problems/intersection-of-two-arrays/ -## 思路 +> 如果哈希值比较少、特别分散、跨度非常大，使用数组就造成空间的极大浪费！ -这道题目，主要要学会使用一种哈希数据结构，unordered_set，这个数据结构可以解决很多类似的问题。 +# 第349题. 两个数组的交集 + +题意：给定两个数组，编写一个函数来计算它们的交集。 + +![349. 两个数组的交集](https://img-blog.csdnimg.cn/20200818193523911.png) + +**说明：** +输出结果中的每个元素一定是唯一的。 +我们可以不考虑输出结果的顺序。 + +# 思路 + +这道题目，主要要学会使用一种哈希数据结构：unordered_set，这个数据结构可以解决很多类似的问题。 注意题目特意说明：**输出结果中的每个元素一定是唯一的，也就是说输出的结果的去重的， 同时可以不考虑输出结果的顺序** -这道题用暴力的解法时间复杂度是O(n^2)，这种解法面试官一定不会满意，那看看使用哈希法进一步优化。 +这道题用暴力的解法时间复杂度是O(n^2)，那来看看使用哈希法进一步优化。 -那么可以发现，貌似用数组做哈希表可以解决这道题目，把nums1的元素，映射到哈希数组的下表上，然后在遍历nums2的时候，判断是否出现过就可以了。 +可以发现，貌似用数组做哈希表可以解决这道题目，把nums1的元素，映射到哈希数组的下表上，然后在遍历nums2的时候，判断是否出现过就可以了。 -但是要注意，使用数据来做哈希的题目，都限制了数值的大小，例如只有小写字母，或者数值大小在[0- 10000] 之内等等。 而这道题目没有限制数值的大小，就无法使用数组来做哈希表了。 +但是要注意，**使用数据来做哈希的题目，都限制了数值的大小，例如[哈希表：可以拿数组当哈希表来用，但哈希值不要太大](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig)题目中只有小写字母，或者数值大小在[0- 10000] 之内等等。** -例如说：如果我的输入样例是这样的， 难道要定义一个2亿大小的数组来做哈希表么， 不同的语言对数组定义的大小都是有限制的， 即使有的语言可以定义这么大的数组，那也是对内存空间造成了非常大的浪费。 +而这道题目没有限制数值的大小，就无法使用数组来做哈希表了。 -此时就要使用另一种结构体了，set ，关于set，C++ 给提供了如下三种可用的数据结构 +**而且如果哈希值比较少、特别分散、跨度非常大，使用数组就造成空间的极大浪费。** + +此时就要使用另一种结构体了，set ，关于set，C++ 给提供了如下三种可用的数据结构： * std::set * std::multiset @@ -25,9 +39,10 @@ https://leetcode-cn.com/problems/intersection-of-two-arrays/ std::set和std::multiset底层实现都是红黑树，std::unordered_set的底层实现是哈希表， 使用unordered_set 读写效率是最高的，并不需要对数据进行排序，而且还不要让数据重复，所以选择unordered_set。 思路如图所示： + ![set哈希法](https://img-blog.csdnimg.cn/2020080918570417.png) -## 代码 +# C++代码 ``` class Solution { public: diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index 3a474d3a..2a72ab56 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -3,7 +3,9 @@ https://leetcode-cn.com/problems/ransom-note/ ## 思路 -这道题题意很清晰，就是用判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成，但是这里需要注意两点1. +这道题目和[242.有效的字母异位词](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig)很像，[242.有效的字母异位词](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig)相当于求 字符串a 和 字符串b 是否可以相互组成 ，而这道题目是求 字符串a能否组成字符串b，而不用管字符串b 能不能组成字符串a。 + +本题判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成，但是这里需要注意两点1. *  第一点“为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表达意思”  这里*说明杂志里面的字母不可重复使用。* @@ -37,13 +39,12 @@ public: }; ``` -这里时间复杂度是比较高的，而且里面还有一个字符串删除也就是erase的操作，也是费时的，当然这段代码也可以过这道题 +这里时间复杂度是比较高的，而且里面还有一个字符串删除也就是erase的操作，也是费时的，当然这段代码也可以过这道题。 -我们想一想优化解法 ## 字典计数法（哈希策略） -因为题目所只有小写字母，那我们可以采用空间换区时间的哈希策略， 用一个长度为26的数组还记录magazine里字母出现的次数，然后再用ransomNote去验证这个数组是否包含了ransomNote所需要的所有字母。 +因为题目所只有小写字母，那可以采用空间换区时间的哈希策略， 用一个长度为26的数组还记录magazine里字母出现的次数，然后再用ransomNote去验证这个数组是否包含了ransomNote所需要的所有字母。 代码如下： ```C++