前言

💫你好，我是辰chen，本文旨在准备考研复试或就业
💫文章题目大多来自于 leetcode，当然也可能来自洛谷或其他刷题平台
💫欢迎大家的关注，我的博客主要关注于考研408以及AIoT的内容
🌟 仅给出C++版代码

以下的几个专栏是本人比较满意的专栏(大部分专栏仍在持续更新)，欢迎大家的关注：

💥ACM-ICPC算法汇总【基础篇】
💥ACM-ICPC算法汇总【提高篇】
💥AIoT(人工智能+物联网)
💥考研
💥CSP认证考试历年题解

数组的度

题目链接：数组的度

C++版AC代码：

class Solution {
public:int findShortestSubArray(vector<int>& nums) {unordered_map<int, int> m;unordered_map<int, int> st, ed;   // 记录该元素第一次出现的坐标以及最后一次出现的坐标for (int i = 0; i < nums.size(); i ++ ) {m[nums[i]] ++;if (!st.count(nums[i])) st[nums[i]] = i;  // 记录第一次出现的坐标ed[nums[i]] = i;                          // 记录最后一次出现的坐标}int md = 0;for (auto i = m.begin(); i != m.end(); i ++ ) {int cnt = i -> second;md = max(md, cnt);        // 找出数组的度}int res = nums.size();for (auto i = m.begin(); i != m.end(); i ++ ) {int num = i -> first, cnt = i -> second;if (cnt == md)res = min(res, ed[num] - st[num] + 1);}return res;}
};

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最短补全词

题目链接：最短补全词

C++版AC代码：

class Solution {
public:string shortestCompletingWord(string licensePlate, vector<string>& words) {unordered_map<char, int> m;            // 统计 licensePlate 中的字符for (int i = 0; i < licensePlate.size(); i ++ ) {char c = licensePlate[i];if (c >= 'A' && c <= 'Z') m[c + 32] ++;   // 根据ASCII大写变小写if (c >= 'a' && c <= 'z') m[c] ++;}string word;                         vector<string> v;                             // 存储补全词for (int i = 0; i < words.size(); i ++ ) {word = words[i];unordered_map<char, int> tmp;             // 统计 word 中的字符for (int j = 0; j < word.size(); j ++ ) {char c = word[j];if (c >= 'A' && c <= 'Z') tmp[c + 32] ++; // 根据ASCII大写变小写if (c >= 'a' && c <= 'z') tmp[c] ++;}bool flag = true;for (auto k = m.begin(); k != m.end(); k ++ ) {char c = k -> first;int cnt = k -> second;if (tmp[c] < cnt) {             // 注意这里是 < flag = false;break;}}if (flag) v.push_back(word);         // 符合补全词的定义}string res = v[0];int minlen = v[0].size();for (int i = 1; i < v.size(); i ++ ) {   // 找最短补全词 if (v[i].size() < minlen) {res = v[i];minlen = v[i].size();}}return res;}
};

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宝石与石头

题目链接：宝石与石头

C++版AC代码：

class Solution {
public:int numJewelsInStones(string jewels, string stones) {unordered_map<char, int> m;for (int i = 0; i < jewels.size(); i ++ ) m[jewels[i]] = 1;int res = 0;for (int i = 0; i <stones.size(); i ++ ) if (m.count(stones[i]))res ++;return res;}
};

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唯一摩尔斯密码词

题目链接：唯一摩尔斯密码词

C++版AC代码：

class Solution {
public:int uniqueMorseRepresentations(vector<string>& words) {unordered_map<string, int> m;unordered_map<char, string> mos;string v[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};for (int i = 0; i < 26; i ++ ) mos['a' + i] = v[i];      // 构建mos表for (int i = 0; i < words.size(); i ++ ) {string word = words[i];string mosi;for (int j = 0; j < word.size(); j ++ ) mosi += mos[word[j]];m[mosi] ++;}return m.size();}
};

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最常见的单词

题目链接：最常见的单词

注意细节处理就好了，关于字符串的操作还需再熟悉

C++版AC代码：

class Solution {
public:unordered_map<string, int> m;string getres() {string res;int maxtimes = 0;for (auto i = m.begin(); i != m.end(); i ++ ) {int times = i -> second;string c = i -> first;if (times > maxtimes) {maxtimes = times;res = c;}}return res;}string mostCommonWord(string paragraph, vector<string>& banned) {for (int i = 0; i < paragraph.size(); i ++ ) {char c = paragraph[i];if ((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z')) {int j = i;while (j < paragraph.size()) {c = paragraph[j];if (c >= 'a' && c <= 'z') j ++;else if (c >= 'A' && c <= 'Z') {paragraph[j] += 32;      j ++;}else break;}string word = paragraph.substr(i, j - i);  // 提取单词m[word] ++;i = j;}else continue;}if (banned.empty()) return getres();for (int i = 0; i <banned.size(); i ++ ) m[banned[i]] = 0;  return getres();}
};

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