#include<iostream>
#include<string>
#include<assert.h>
#include <vector>
#include <algorithm>
using namespace std;
//数字转字符串
string InttoChar(int num) {
if (num == 0)
return "0";
string res;
while (num != 0) {
int tmp = num % 10;
char ctmp = tmp + '0';
res.push_back(ctmp);
num /= 10;
}
reverse(res.begin(), res.end());
return res;
}
//字符串逆置
string reversestring(string s1) {
if (s1 == "")
return s1;
string s2 = "";
for (int i = s1.length() - 1; i >= 0; i--) {
s2 += s1[i];
}
return s2;
}
//折半法逆置字符串
char* revetsestring(char* s1) {
if (s1 == NULL)
return s1;
int len = strlen(s1);
char* p1, *p2;
p1 = s1, p2 = s1 + len - 1;
char tmp;
while (p1 != p2) {
tmp = *p1;
*p1 = *p2;
*p2 = tmp;
p1++;
p2--;
}
return s1;
}
//异或法逆置字符串
char* revetsestr(char* str) {
if (str == NULL)
return str;
char* f = str, *b = str + strlen(str) - 1;
while (f<b) {
*f ^= *b;
*b ^= *f;
*f ^= *b;
f++;
b--;
}
return str;
}
//字符串逆置单词内部顺序不变
char* ReverseWords(char* s1) {
int len = strlen(s1);
if (s1 == NULL || len == 1)
return s1;
int i = len - 1, j = len; //ij记录单词的位置
int t = 0; //暂存已经录入单词的位置
char* res = new char[len + 1]; //len不包含'\0'要申请len+1个
int k = 0; //新字符串的位置
while (i>0) {
while (s1[i] != ' '&&i != 0)
i--;
t = i;
if (i != 0)
{
i++;
while (i<j) {
res[k++] = s1[i++];
}
res[k++] = ' ';
j = t;
i = t - 1;
}
else {
while (i<j)
{
res[k++] = s1[i++];
}
i = t;
}
}
res[len] = '\0';
return res;
}
//分别实现strcpy,strcat,strcmp,strstr
//strcpy需要一个指针记录开始的位置
char* stringcpy(char* des, char* src) {
if (src == NULL)
return src;
int len = strlen(src);
des = new char[len + 1];
char* res = des;
while (*src != '\0')
*des++ = *src++;
*des = '\0';
return res;
}
//strcat,strcat需要des保证可以容纳src，我们这里不需要
char* stringcat(char* des, char*src) {
if (src == NULL)
return des;
int len1 = strlen(des);
int len2 = strlen(src);
char* newstr = new char[len1 + len2 + 1];
char* res = newstr;
while (*des != '\0')
*newstr++ = *des++;
while (*src != '\0')
*newstr++ = *src++;
*newstr = '\0';
return res;
}
//strcmp,使用assert要包含其头文件assert.h不满足assert条件会报错
int stringcmp(char* s1, char* s2) {
assert((s1 != NULL) && (s2 != NULL));
while (*s1&&*s2 && (*s1 == *s2)) {
s1++;
s2++;
}
return *s1 - *s2;
}
//strstr leetcode有原题，做过
bool samestr(char* s1, char* s2) {
if (strlen(s1) != strlen(s2))
return false;
while (*s1) {
if (*s1 == *s2) {
s1++;
s2++;
}
else return false;
}
return true;
}
char* stringstr(char* s1, char* s2) {
if (s1 == NULL || s1 == NULL)
return NULL;
int len1 = strlen(s1);
int len2 = strlen(s2);
if (len2>len1)
return NULL;
if (len1 == len2) {
if (samestr(s1, s2))
return s1;
else return NULL;
}
for (int i = 0; i<len1;) {
char* tmp = NULL;
for (int j = 0; j<len2; j++)
tmp += s1[i + j];
tmp += '\0';
if (samestr(tmp, s2))
return s1 + i;
else i++;
}
return NULL;
}
//最长公共子串
string MaxSubstring(string& s1, string& s2) {
if (s1 == "" || s2 == "")
return "";
int len1 = s1.length();
int len2 = s2.length();
int arraystr[100][100] = { 0 };//then the max length of string is 1000
for (int i = 0; i<len1; i++) {
for (int j = 0; j<len2; j++) {
if (s1[i] == s2[j])
{
if (i == 0 || j == 0)
arraystr[i][j] == 1;
else arraystr[i][j] = arraystr[i - 1][j - 1] + 1;
}
else arraystr[i][j] = 0;
}
}
int maxlen = 0;
int maxindex = 0;
for (int i = len1 - 1; i >= 0; i--) {
for (int j = len2 - 1; j >= 0; j--) {
if (arraystr[i][j]>maxlen) {
maxlen = arraystr[i][j];
maxindex = j;
}
}
}
string res = "";
for (int k = maxindex - maxlen + 1; k <= maxindex; k++)
res += s2[k];
return res;
}
//最长公共子序列长度
int Maxsubsequencelength(string& s1, string& s2) {
if (s1 == "" || s2 == "")
return 0;
int arraystr[1010][1010] = { { 0,0 } };
int len1 = s1.length();
int len2 = s2.length();
for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (s1[i - 1] == s2[j - 1])
arraystr[i][j] = arraystr[i - 1][j - 1] + 1;
else arraystr[i][j] = max(arraystr[i - 1][j], arraystr[i][j - 1]);
}
}
return arraystr[len1][len2];
}
string Maxsubsequence(string& s1, string& s2) { //输入时字符串长的那个为第一参数
string res = "";
if (s1 == "" || s2 == "")
return res;
int arraystr[100][100] = { { 0,0 } };
int len1 = s1.length();
int len2 = s2.length();
for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (s1[i - 1] == s2[j - 1])
arraystr[i][j] = arraystr[i - 1][j - 1] + 1;
else arraystr[i][j] = max(arraystr[i - 1][j], arraystr[i][j - 1]);
}
}
for (int i = len1; i >= 0;) {
for (int j = len2; j >= 0;) {
if (arraystr[i][j]>max(max(arraystr[i - 1][j], arraystr[i][j - 1]), arraystr[i - 1][j - 1])) {
res = s1[i] + res;
i--;
j--;
}
else {
if (arraystr[i][j] == arraystr[i - 1][j - 1]) {
i--;
j--;
}
else if (arraystr[i][j] == arraystr[i - 1][j])
i--;
else if (arraystr[i][j] == arraystr[i][j - 1])
j--;
}
}
}
return res;
}
//最后一个单词的长度
int lastword(string s1) {
if (s1 == "")
return 0;
int len = s1.length();
int i = len - 1;
int res = 0;
while (i >= 0 && s1[i] == ' ') {
i--;
}
while (i >= 0 && s1[i] != ' ') {
i--;
res++;
}
return res;
}
int lastwordlen(string s1) {
getline(cin, s1);
if (s1 == "")
cout << 0 << endl;
return s1.length() - s1.rfind(' ') - 1;
}
//回文串
bool isPalindrome(string s1) {
string s2 = s1;
reverse(s2.begin(), s2.end());
if (s1 == s2)
return true;
else return false;
}
//字符串大数加法
string stringadd(string& s1, string& s2) {
int len1 = s1.length();
int len2 = s2.length();
if (len1 == 0)
return s2;
if (len2 == 0)
return s1;
int mlen = max(len1, len2);
int* arr1 = new int[mlen];//把字符串数字转换成数组存储，在进行加和然后转成字符串
int* arr2 = new int[mlen];
int index = mlen - 1;
for (int i = len1 - 1, j = len2 - 1; i >= 0 || j >= 0; index--) {
if (i >= 0)
arr1[index] = s1[i--] - '0';
else arr1[index] = 0;
if (j >= 0)
arr2[index] = s2[j--] - '0';
else arr2[index] = 0;
}
int* res = new int[mlen + 1];
int carr = 0;
int k = mlen - 1, l = mlen;
while (k >= 0) {
int tmp = arr1[k] + arr2[k] + carr;
carr = tmp / 10;
res[l--] = tmp % 10;
k--;
}
if (carr != 0)
res[0] = carr;
else res[0] = 0;
string ret = "";
for (int i = 0; i <= mlen; i++) {
if (res[i] == 0)
continue;
char c = '0' + res[i];
ret = ret + c;
}
return ret;
}
//字符串大数乘法
char* MultiString(char* str1, char* str2) {
int len1 = strlen(str1), len2 = strlen(str2);
if (len2>len1) { //len1,str1对应较长的字符串
char* temp = str1;
int temp1 = len1;
str1 = str2;
str2 = temp;
len1 = len2;
len2 = temp1;
}
char* ret = new char[len1 + len2];
memset(ret, '0', sizeof(char)*(len1 + len2 - 1)); //初始化字符串
ret[len1 + len2 - 1] = '\0';
if (len1 == 0 || len2 == 0)
return ret;
int temp1 = 0, temp2 = 0;
for (int i = len2 - 1; i >= 0; i--) {
int carry = 0;
for (int j = len1 - 1; j >= 0; j--) {
int sum = ret[i + j] - '0';
if (carry>0)
sum += carry;
temp1 = str1[j] - '0';
temp2 = str2[i] - '0';
sum += temp1*temp2;
if (sum >= 10) {
carry = sum / 10;
}
else carry = 0;
ret[i + j] = sum % 10 + '0'; //规格化处理
if (j == 0 && carry>0) //最高位以后还有可能进位
ret[i - 1] += carry;
}
cout << i << " " << ret << endl;
}
char* p = ret;
while (*p == '0') //如果前面有0，从第一位不是0开始输出
p++;
cout << p << endl;
return p;
}
int main() {
int num = 1234569;
string res = InttoChar(num);
cout << res << endl;
char* test = "guofei is a student in SEU!";
string test1 = "helloworld";
char* test2 = "happyforever";
cout << reversestring(test2) << endl;
cout << reversestring(test1) << endl;
cout << ReverseWords(test) << endl;
char* s1, *s2 = "cpy this string";
cout << stringcpy(s2, s2) << endl;
cout << stringcmp(test, test2) << endl;
char* s3 = "auti";
char* s4 = "beautiful girl";
if (stringstr(s3, s4))
cout << "fei zi chuan" << endl;
else cout << "zichuan" << endl; //his problem is really tricky, i try to cout NULL and i fix bug just for this little error makes me very wordless
string s5 = "ade";
string s6 = "abcde";
string res1 = MaxSubstring(s5, s6);
cout << res1 << endl;
cout << Maxsubsequence(s6, s5) << endl;
string s7 = "accba";
cout << isPalindrome(s7) << endl;
return 0;
}
