从一个长串中找出没有相同字符的最长子串。
int find_length(std::string temp,char c)
{
printf(“temp %s,%c.\n”,temp.c_str(),c);
int length = temp.length();
for(int i = 0;i<temp.length();i++)
{
if(temp[i] == c)
{
return temp.length();
if(temp.length() – i > i)
{
return temp.length() – i;
}
else
{
return i;
}
}
}
return -1;
}
int GetLongestSignelString(std::string value)
{
std::string temp;
int max_count = 0;
for(int i = 0;i<value.length();i++)
for(int j = i + 1;j<value.length();j++)
{
temp = value.substr(i,j-i);
int found_len = find_length(temp,value[j]);
printf(” found_len %d.############ i %d j %d len %d\n”,found_len,i,j,temp.length());
if(found_len > 0)
{
if(found_len > max_count)
{
max_count = found_len;
}
printf(“break 11.max_count %d\n”,max_count);
break;
}
else
{
if(j == value.length() – 1)
{
if(temp.length() + 1 > max_count)
{
max_count = temp.length() + 1;
}
printf(“break here.max_count %d\n”,max_count);
break;
}
printf(“break 22.\n”);
}
}
return max_count;
}
int main()
{
std::string str(“accbjhbsaalkbsdwie12324326492!@#%$&^*VMYHG”);
printf(“longest %d.\n”,GetLongestSignelString(str));
return 0;
}
最近优化了一个高效一些的版本。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <iostream>
int findLast(char* a, int len, char c)
{
if (len == 0) {
return 0;
}
for (int i = len – 1; i >= 0;i –)
if (a[i] == c) {
return i + 1;
}
return 0;
}
std::string SingleStr(const std::string& str)
{
int validlen = 0;
std::string result;
char *bufstr = (char*)malloc(str.length());
for (int i = 0; i<str.length(); i++)
{
int find = findLast(bufstr, validlen, str[i]);
if (find != 0) {
if (find > result.length()) {
result.assign(bufstr, find);
}
memmove(bufstr, bufstr + find, validlen – find);
validlen -= find;
}
bufstr[validlen ++] = str[i];
if (validlen > result.length()) {
result.assign(bufstr, validlen);
}
}
if (validlen > result.length()) {
result.assign(bufstr, validlen);
}
free(bufstr);
return result;
}
int main()
{
std::string a(“120135435”);
std::cout<<SingleStr(a)<<std::endl;
}
以上使用的是蛮力比的过程输出，不过效率不高，比较耗时