A - identical test
Traverse from the beginning, output subscripts when encountering the same characters, and traverse to a relatively short string.
Space complexity O(1), time complexity O(n)
code:
#include <bits/stdc++.h>
using namespace std;
int main(){
string str1;
string str2;
getline(cin,str1);
getline(cin,str2);
for(int i=0;i<min(str1.size(),str2.size());i++){
if(str1[i]==str2[i]){
cout<<i+1<<" ";
}
}
return 0;
}B - initial capital
Traverse the string to determine whether the first letter and the previous letter with a space are lowercase. If they are lowercase, they will be converted to uppercase
Space complexity O(1), time complexity O(n)
code:
#include <bits/stdc++.h>
using namespace std;
int main(){
string str;
getline(cin,str);
for(int i=0;i<str.size();i++){
if(i==0&&str[i]>='a'){
str[i]-=32;
}
if(i>0&&str[i-1]==' '&&str[i]>='a'){
str[i]-=32;
}
}
cout<<str;
return 0;
}C - case conversion
Traverse the string and convert lowercase letters to uppercase letters
Space complexity O(1), time complexity O(n)
#include <bits/stdc++.h>
using namespace std;
int main(){
string str;
while(cin>>str){
for(int i=0;i<str.size();i++){
if(str[i]>='a'&&str[i]<='z'){
str[i]-=32;
}
}
cout<<str<<endl;
}
return 0;
}D - Digital inversion
First determine whether it is a positive or negative number, and then use the reverse function and stoi function to reverse the string and convert the string to int type
Space complexity O(1), time complexity O(n)
#include <bits/stdc++.h>
using namespace std;
int main(){
string num;
cin>>num;
int a;
if(num[0]=='-'){
reverse(num.begin()+1,num.end());
a=stoi(num);
}
else{
reverse(num.begin(),num.end());
a=stoi(num);
}
cout<<a;
return 0;
} E - delete word suffix
Use the find function to find, and then use the substr function to intercept
Space complexity O(1), time complexity O(n2)
#include <iostream>
#include <string>
using namespace std;
string s;
int main(){
cin >> s;
if (s.find("er", s.size()-2) != -1)
s = s.substr(0, s.size()-2);
else if (s.find("ly", s.size()-2) != -1)
s = s.substr(0, s.size()-2);
else if (s.find("ing", s.size()-3) != -1)
s = s.substr(0, s.size()-3);
cout << s << endl;
return 0;
}F - judge whether the string is palindrome
Use the reverse function to reverse, and then judge whether it is the same after inversion and before inversion
Spatial complexity O(1), temporal complexity O(n)
#include <bits/stdc++.h>
using namespace std;
int main(){
string str;
cin>>str;
string temp=str;
reverse(str.begin(),str.end());
if(temp==str){
cout<<"yes";
}
else{
cout<<"no";
}
return 0;
} G - basic data structure - stack (1)
Traverse the string, enter the stack when encountering {, (, [and exit the stack when encountering},),]
Space complexity O(n), time complexity O(n)
#include <bits/stdc++.h>
using namespace std;
int main(){
string str;
while(getline(cin,str)){
stack<char> s;
for(int i=0;i<str.size();i++){
if(str[i]=='('||str[i]=='['||str[i]=='{')
s.push(str[i]);
else if(str[i]==')'||str[i]==']'||str[i]=='}'){
if(s.empty())
{
s.push(str[i]);
break;
}
if(str[i]==')'&&s.top()=='(') s.pop();
else if(str[i]==']'&&s.top()=='[') s.pop();
else if(str[i]=='}'&&s.top()=='{') s.pop();
}
}
if(s.size()) cout<<"no"<<endl;
else cout<<"yes"<<endl;
}
return 0;
}H - dictionary order
For direct comparison, the comparison function of string is overloaded in C + +
Space complexity O(1), time complexity O(n)
#include <bits/stdc++.h>
using namespace std;
int main(){
string str1,str2;
getline(cin,str1);
getline(cin,str2);
if(str1<str2){
cout<<"YES";
}
else cout<<"NO";
return 0;
}I - verify substring
Use the find function to find
Space complexity O(1), time complexity O(n2)
#include <bits/stdc++.h>
using namespace std;
int main(){
string str1;
string str2;
cin>>str1>>str2;
if(str2.find(str1)!=string::npos) cout<<str1<<" is substring of "<<str2<<endl;
else if(str1.find(str2)!=string::npos) cout<<str2<<" is substring of "<<str1<<endl;
else cout<<"No substring"<<endl;
return 0;
}J - substring lookup
kmp algorithm, pay attention to overlap
Space complexity O(n), time complexity O(n+m)
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define maxn 1000005
using namespace std;
char text[maxn], patten[maxn];
int net[maxn];
int main() {
scanf("%s%s",text,patten);
int la = strlen(text);
int lb = strlen(patten);
memset(net, 0, sizeof(net));
for (int i = 1; i < lb; ++i) {
int j = i;
while (j > 0) {
j = net[j];
if (patten[j] == patten[i]) {
net[i + 1] = j + 1;
break;
}
}
}
int cnt = 0;
for (int i = 0, j = 0; i < la; ++i) {
if (j < lb&&patten[j] == text[i]) {
j++;
}
else {
while (j > 0) {
j = net[j];
if (text[i] == patten[j]) {
j++;
break;
}
}
}
if (j == lb) cnt++;
}
printf("%d\n", cnt);
return 0;
}K - cut cloth strip
kmp, no overlap
Space complexity O(n), time complexity O(n+m)
#include <bits/stdc++.h>
using namespace std;
const int N=1010;
int c[N];
void getc(string b){
int j=0;
c[j]=0;
for(int i=1;i<b.length();i++){
while(b[i]!=b[j]&&j>0){
j=c[j-1];
}
if(b[i]==b[j]){
j+=1;
}
c[i]=j;
}
}
int KMP(string a,string b){
int cnt=0;
int j=0;
memset(c,0,sizeof(c));
getc(b);
for(int i=0;i<a.length();i++){
if(a[i]!=b[j]&&j>0){
j=c[j-1];
}
if(a[i]==b[j]){
j+=1;
}
if(j==b.length()){
j=0;
cnt++;
}
}
return cnt;
}
int main(){
string a,b;
while(cin>>a)
{
if(a=="#") break;
cin>>b;
cout<<KMP(a,b)<<endl;
}
return 0;
}L - longest palindrome substring
Using dynamic programming to find
Space complexity O(n), time complexity O(n2)
#include <bits/stdc++.h>
using namespace std;
int main(){
string str;
cin>>str;
if (str.size()<2) {
return str.size();
}
else{
bool dp[str.size()][str.size()];
int maxlen=1;
int k=0;
for(int i=0;i<str.size();i++){
dp[i][i]=true;
}
for (int j = 1; j < str.size(); j++) {
for (int i = 0; i < j; i++) {
if (str[i] != str[j]) {
dp[i][j] = false;
}
else {
if (j - i < 3) {
dp[i][j] = true;
}
else {
dp[i][j] = dp[i + 1][j - 1];
}
}
if (dp[i][j] && j - i + 1 > maxlen) {
maxlen = j - i + 1;
k = i;
}
}
}
cout<<maxlen;
}
return 0;
}