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string_cycle_shift.cpp
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string_cycle_shift.cpp
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//http://m.newsmth.net/article/CoderInterview/1451?p=1
//1,在一个字符串中找到第一个只出现一次的字符。
//这题后来在网上一搜全是答案,不过都好像只是hash。
//这个当时下意识说了第一遍哈希,第二遍就能找到。
//面试官继续问,如果进行一遍扫描呢,我就卡住了。
//
//
//2,两个大文本文件中相同的行输出出来。求交集。
//
//3.右移一个字符串,比如:abcdefg-> fgabcde
#include <stdio.h>
#include <string>
#include <algorithm>
// copy from libcxx __rotate_gcd
void cstring_shift(char pChar[], int size, int shiftcycle) {
int exchangePos = size - shiftcycle - 1;
int exchangeSize = exchangePos+1;
if (exchangeSize == shiftcycle) {
for (int i = 0; i < exchangeSize; ++i)
{
char tmpchar = pChar[i];
pChar[i] = pChar[exchangeSize + i];
pChar[exchangeSize + i ] = tmpchar;
}
return;
}
int p = exchangeSize;
int q = shiftcycle;
do {
int t = p % q;
p = q;
q = t;
} while(q);
for (char * pp = pChar + p; pp != pChar; ) {
int t = *--pp;
char *pp1 = pp;
char *pp2 = pp1 + exchangeSize;
do {
*pp1 = *pp2;
pp1 = pp2;
const int d = pChar + size - pp2;
if (exchangeSize < d)
{
pp2 += exchangeSize;
}
else
{
pp2 = pChar + (exchangeSize - d);
}
} while(pp2 != pp);
*pp1 = t;
}
}
void string_shift(std::string& input, int shiftcycle) {
std::string::iterator midIter = input.begin() + input.size() - shiftcycle;
std::rotate(input.begin(), midIter, input.end() );
printf("after shift :%s\n", input.c_str() );
}
int main() {
//char* p = "abcdefg";
//std::rotate(0, 2, sizeof(p));
std::string str = "abcdefg";
string_shift(str, 2);
//std::rotate(str.begin(), str.begin() + 2, str.end());
//http://www.cplusplus.com/reference/algorithm/rotate/
//Rotates the order of the elements in the range [first,last), in such a way that the element pointed by middle becomes the new first element.
str = "abcdefg";
std::rotate(str.begin(), str.begin() + 5, str.end());
printf("%s\n", str.c_str());
char pStr[] = "abcdefg";
cstring_shift(pStr, 7, 2);
printf("%s\n", pStr);
char pStr2[] = "abcdefgh";
cstring_shift(pStr2, 8, 4);
printf("%s\n", pStr2 );
}
#if 0
//73楼|CrTn|2013-05-06 22:41:47|只看此ID
//从屁股往前扫一遍不就可以了么
#include <iostream>
#include <unordered_map>
#include <type_traits>
#include <string>
#include <vector>
using namespace std;
template <typename T>
T firstUnique(const T& beg, const T& end) {
using keyType = typename remove_const<
typename remove_reference<decltype(*beg)
>::type>::type;
unordered_map<keyType, size_t> stat;
auto it = beg;
auto res = beg;
while(it != end) {
const auto &ref = *it;
++stat[ref];
if(stat[ref] == 1)
res = it;
++it;
}
const auto &ref = *res;
if(stat[ref] == 1)
return res;
else {
return end;
}
}
int main()
{
string str = "afdafgag";
auto cit = firstUnique(str.crbegin(), str.crend());
if(cit != str.crend())
cout << *cit << endl;
else
cout << "No Unique" << endl;
vector<int> ivec{1,2,3,2,3,4,5,1,9,3,3,2,5,6};
auto iit = firstUnique(ivec.crbegin(), ivec.crend());
if(iit != ivec.crend())
cout << *iit << endl;
else
cout << "No Unique" << endl;
return 0;
}
#endif
#if 0
linkedhashtable
27楼|bxhsix|2012-03-30 15:08:11|
第一道应该是hash表+双向链表,简单写了一下,应该可以一遍解决问题
#include <iostream>
#include <map>
using namespace std;
struct pointer
{
int prev_chr;
int next_chr;
};
int main(int argc, char *argv[])
{
if (argc < 2)
{
cout << "Usage: " << argv[0] << " <InputString>" << endl;
return -1;
}
char *str = argv[1];
map<char, pointer> hash;
int first_uniq_chr = -1;
int last_chr = -1;
for (int i=0; str[i]!='\0'; ++i)
{
if (first_uniq_chr == -1)
{
if (hash.find(str[i]) == hash.end())
{
first_uniq_chr = i;
last_chr = i;
hash[str[i]] = (pointer){-1, -1};
}
continue;
}
if (str[i] == str[first_uniq_chr])
{
first_uniq_chr = hash[str[first_uniq_chr]].next_chr;
}
else
{
if (hash.find(str[i]) == hash.end())
{
hash[str[i]] = (pointer){last_chr, -1};
hash[str[last_chr]].next_chr = i;
last_chr = i;
}
else
{
if (hash[str[i]].prev_chr != -2)
{
hash[str[hash[str[i]].prev_chr]].next_chr = hash[str[i]].next_chr;
hash[str[hash[str[i]].next_chr]].prev_chr = hash[str[i]].prev_chr;
if (str[i] == str[last_chr])
{
last_chr = hash[str[last_chr]].prev_chr;
}
hash[str[i]].prev_chr = -2;
}
}
}
}
cout << first_uniq_chr << ": " << str[first_uniq_chr] << endl;
return 0;
}
#endif