💫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
题目链接:验证外星语词典
C++版AC代码:
class Solution {
public:
bool isAlienSorted(vector<string>& words, string order) {
unordered_map<char, char> m;
// 注意这里相当于重新映射成 a b c d e f g ...
for (int i = 0; i < order.size(); i ++ ) m[order[i]] = (char)(i + 'a');
int n = words.size();
vector<string> tmp, exch;
for (int i = 0; i < n; i ++ ) {
string word = words[i], numstr;
for (int j = 0; j < word.size(); j ++ )
numstr += m[word[j]]; // 将原字符串映射为按照外星语字典序映射的字符串,即可排序
tmp.push_back(numstr), exch.push_back(numstr); // 分别存到tmp(要排序),exch(对比串)中
}
sort(tmp.begin(), tmp.end()); // 对tmp进行排序
bool flag = true;
for (int i = 0; i < n; i ++ )
if (tmp[i] != exch[i]) { // 有变化即不是按照新字典序有序排列
flag = false;
break;
}
return flag;
}
};
C++版AC代码:
class Solution {
public:
int repeatedNTimes(vector<int>& nums) {
unordered_map<int, int> m;
int res;
for (int i = 0; i < nums.size(); i ++ ) {
m[nums[i]] ++;
if (m[nums[i]] >= 2) { // 因为有n+1个不同的值,所以当一个元素出现2次的时候就是目标值
res = nums[i];
break;
}
}
return res;
}
};
题目链接:找到小镇的法官
C++版AC代码:
class Solution {
public:
int findJudge(int n, vector<vector<int>>& trust) {
if (trust.empty() && n == 1) return 1;
int judge = -1;
unordered_map<int, int> m1;
unordered_map<int, bool> m2;
for (int i = 0; i < trust.size(); i ++ ) {
int fs = trust[i][0], sd = trust[i][1];
m1[sd] ++, m2[fs] = false; // 信任sd的人数+1, fs不可能是法官
}
for (auto i = m1.begin(); i != m1.end(); i ++ ) {
int guy = i -> first, num = i -> second;
if ((num == n - 1) && !m2.count(guy)) {
judge = guy;
break;
}
}
return judge;
}
};
题目链接:查找共用字符
C++版AC代码:
class Solution {
public:
vector<string> commonChars(vector<string>& words) {
unordered_map<char, int> m; // 用来记录共用字符,一开始把words[0]存进去
for (int i = 0; i < words[0].size(); i ++ ) m[words[0][i]] ++;
for (int i = 0; i < words.size(); i ++ ) {
string word = words[i];
unordered_map<char, int> tmp; // 存储当前字符串的信息,用来和m做对比
for (int j = 0; j < word.size(); j ++ )
tmp[word[j]] ++;
for (auto j = m.begin(); j != m.end(); j ++ ) {
char s = j -> first;
int cnt = j -> second;
if (!tmp.count(s)) m[s] = 0; // tmp中没有这个字符,即不是共用字符,删掉
else m[s] = min(cnt, tmp[s]); // 如果是共用字符则取出现次数为两个串中的最小值
}
}
vector<string> res;
for (auto i = m.begin(); i != m.end(); i ++ ) {
int cnt = i -> second;
while (cnt -- ) res.push_back(string(1, i -> first));
// string(1, c) : 把字符c变成字符串格式
}
return res;
}
};
题目链接:数组的相对排序
C++版AC代码:
class Solution {
public:
vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
unordered_map<int, int> m;
for (int i = 0; i < arr1.size(); i ++ ) m[arr1[i]] ++;
vector<int> res, tmp;
for (int i = 0; i < arr2.size(); i ++ ) {
auto it = m.find(arr2[i]);
int k = it -> second;
while (k -- ) res.push_back(arr2[i]); // 按照arr2中的顺序把arr1中的元素存入res
it -> second = 0; // 标记成已经存储好
}
for (auto i = m.begin(); i != m.end(); i ++ ) { // 把arr2中没有的元素暂存到tmp中
int k = i -> second;
while (k -- ) tmp.push_back(i -> first);
}
sort(tmp.begin(), tmp.end()); // arr2中没有的元素排序
for (int i = 0; i < tmp.size(); i ++ ) res.push_back(tmp[i]);
return res;
}
};
题目链接:分发饼干
C++版AC代码:
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(), g.end()), sort(s.begin(), s.end());
int res = 0;
for (int i = 0, j = 0; i < g.size() && j < s.size(); j ++ )
if (g[i] <= s[j]) i ++, res ++;
return res;
}
};
题目链接:分发糖果
C++版AC代码:
class Solution {
public:
int candy(vector<int>& ratings) {
int n = ratings.size();
vector<int> v(n, 1); // 开一个长度为n的vector并附初值为1
for (int i = 1; i < n; i ++ ) // 从头往后遍历
if (ratings[i] > ratings[i - 1]) // 右边比左边大就让右边+1
v[i] = v[i - 1] + 1;
for (int i = n - 1; i > 0; i -- )
if (ratings[i] < ratings[i - 1]) // 左边比右边大且左边的糖果数≤右边的糖果数就更新+1
v[i - 1] = max(v[i - 1], v[i] + 1);
return accumulate(v.begin(), v.end(), 0);
}
};
题目链接:区间选点(AcWing)
C++版AC代码:
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
struct St{
int l, r;
bool operator < (const St w) const { // 按照右端点进行排序
return r < w.r;
}
}st[N];
int main()
{
int n;
cin >> n;
for (int i = 0; i < n; i ++ ) {
int l, r;
cin >> l >> r;
st[i] = {l, r};
}
sort(st, st + n);
int ed = -1e9 - 1, res = 0;
for (int i = 0; i < n; i ++ ) {
if (st[i].l > ed) { // 新区间的左端点大于当前区间的右端点,证明这是一个新的区间
res ++; // 区间数 + 1
ed = st[i].r; // 更新为新的右端点
}
}
cout << res;
return 0;
}
题目链接:最大不相交区间数量(AcWing)
C++版AC代码:
同上
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int N = 100010;
struct St {
int l, r;
bool operator < (const St w) const {
return r < w.r;
}
}st[N];
int main()
{
int n;
cin >> n;
for (int i = 0; i < n; i ++ ) {
int l, r;
cin >> l >> r;
st[i] = {l, r};
}
sort(st, st + n);
int res = 0, ed = -1e9 - 1;
for (int i = 0; i < n; i ++ )
if (st[i].l > ed) {
res ++;
ed = st[i].r;
}
cout << res;
return 0;
}
题目链接:无重叠区间
C++版AC代码:
class Solution {
public:
// 问题等价于求最多的无重叠区间
// 贪心思路:按照右端点进行排序,想要最多的无重叠区域,就是在不重叠的时候选择最小的右端点
static bool cmp(const vector<int> &a, const vector<int> &b) {
return a[1] < b[1];
}
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), cmp);
int res = 0, r = -5 * 1e4 - 1;
for (int i = 0; i < intervals.size(); i ++ ) {
int st = intervals[i][0], ed = intervals[i][1];
if (st >= r) { // 根据例子所述,区间是不包含两个端点的,所以相等的时候也需要更新
res ++;
r = ed;
}
}
return intervals.size() - res;
}
};
定义结构体的写法:
struct St {
int l, r;
bool operator < (const St w) const {
return r < w.r;
}
}st[N];
// 调用无需传入第三个参数:
sort(st, st + n);
关于外部定义的写法:
static bool cmp(const vector<int> &a, const vector<int> &b) {
return a[1] < b[1];
}
// 调用需要传入第三个参数:
sort(intervals.begin (), intervals.end(), cmp);