// CPP Program to find all the common characters
// in n strings
#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;
void commonCharacters(string str[], int n)
{
// primary array for common characters
// we assume all characters are seen before.
bool prim[MAX_CHAR] = {true};
//memset(prim, true, sizeof(prim));
// for each string
for (int i = 0; i < n; i++) {
// secondary array for common characters
// Initially marked false
bool sec[MAX_CHAR] = { false };
// for every character of ith string
for (int j = 0; str[i][j]; j++) {
// if character is present in all
// strings before, mark it.
if (prim[str[i][j] - 'a'])
sec[str[i][j] - 'a'] = true;
}
// copy whole secondary array into primary
//memcpy(prim, sec, MAX_CHAR);
for(int k=0;k<n;k++)
prim[k] = sec[k];
}
// displaying common characters
for (int i = 0; i < 26; i++)
if (prim[i])
printf("%c ", i + 'a');
}
// Driver's Code
int main()
{
string str[] = { "geeksforgeeks",
"gemkstones",
"acknowledges",
"aguelikes" };
int n = sizeof(str)/sizeof(str[0]);
commonCharacters(str, n);
return 0;
}
我们使用尺寸26的两个哈希阵列(对于A-Z,其中0是一个,且z是25)。方法很简单,如果我们看到一个字符之前,我们将纪念它,如果我们没有那么忽略字符,因为它不是一个常见的一种。为什么会发生这种代码没有给出所需的输出?而如果我用memset(prim,true,sizeof(prim))
代替bool prim[MAX_CHAR] = {true};
的初始化和memcpy(prim,sec,MAX_CHAR)
代替for(int k=0;k<n;k++) prim[k] = sec[k];
的用于复制布尔数组秒[]在一本正经[]它工作得很好。
警告
bool prim[MAX_CHAR] = {true};
不等同于memset(prim, true, sizeof(prim));
MAX_CHAR
是26,你只给1个值与{true}
,所以prim[0]
初始化与真实,所有的25个条目0(假)初始化。真不使用到该阵列的端部。
但
bool prim[MAX_CHAR] = {true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true, true }
初始化26项(如果我算好)
当然memcpy(prim,sec,MAX_CHAR)
和for(int k=0;k<n;k++) prim[k] = sec[k];
是不等效的,因为n是串(4)的数量和不重视MAX_CHAR
(26)
执行与您的代码:
pi@raspberrypi:/tmp $ ./a.out
pi@raspberrypi:/tmp $
执行与memset的或{}
26属实,并通过for(int k=0;k<n;k++)
更换for(int k=0; k<MAX_CHAR; k++)
:
pi@raspberrypi:/tmp $ ./a.out
e g k s pi@raspberrypi:/tmp $
从弗朗索瓦·安德里厄的提案(在下面一个移除的话),以除去该问题有关的整洁的初始化:扭转拘谨的布尔值,所以
void commonCharacters(string str[], int n)
{
// primary array for common characters
// we assume all characters are seen before.
// (false means seen before, reverse logic)
bool prim[MAX_CHAR] = {false};
// for each string
for (int i = 0; i < n; i++) {
// secondary array for common characters
// Initially marked false (standard logic)
bool sec[MAX_CHAR] = { false };
// for every character of ith string
for (int j = 0; str[i][j]; j++) {
// if character is present in all
// strings before, mark it.
if (!prim[str[i][j] - 'a'])
sec[str[i][j] - 'a'] = true;
}
// copy negation of whole secondary array into primary
for(int k=0; k<MAX_CHAR; k++)
prim[k] = !sec[k];
}
// displaying common characters
for (int i = 0; i < 26; i++)
if (!prim[i])
printf("%c ", i + 'a');
}
执行:
pi@raspberrypi:/tmp $ ./a.out
e g k s pi@raspberrypi:/tmp $
但对于整洁的反向逻辑和秒的标准逻辑可以是混乱