原文:
给定两个字符串，按照顺序打印所有常用字符。如果没有常用字母，打印-1。所有字母都是小写的。
示例:
input :
string1 : geeks
string2 : forgeeks
output : eegks
explanation: the letters that are common between
the two strings are e(2 times), k(1 time) and
s(1 time).
hence the lexicographical output is "eegks"
input :
string1 : hhhhhello
string2 : gfghhmh
output : hhh
想法是使用字符计数数组。 1)统计第一个和第二个字符串中从“a”到“z”的所有字符的出现次数。将这些计数存储在两个数组 a1[]和 a2[]中。 2)遍历 a1[]和 a2。对于每个索引 i，打印字符' a' i 的次数等于 min(a1[i]，a2[i])。
下面是上述步骤的实现。
c
// c program to print common characters
// of two strings in alphabetical order
#include
using namespace std;
int main()
{
string s1 = "geeksforgeeks";
string s2 = "practiceforgeeks";
// to store the count of
// letters in the first string
int a1[26] = {0};
// to store the count of
// letters in the second string
int a2[26] = {0};
int i , j;
char ch;
char ch1 = 'a';
int k = (int)ch1, m;
// for each letter present, increment the count
for(i = 0 ; i < s1.length() ; i )
{
a1[(int)s1[i] - k] ;
}
for(i = 0 ; i < s2.length() ; i )
{
a2[(int)s2[i] - k] ;
}
for(i = 0 ; i < 26 ; i )
{
// the if condition guarantees that
// the element is common, that is,
// a1[i] and a2[i] are both non zero
// means that the letter has occurred
// at least once in both the strings
if (a1[i] != 0 and a2[i] != 0)
{
// print the letter for a number
// of times that is the minimum
// of its count in s1 and s2
for(j = 0 ; j < min(a1[i] , a2[i]) ; j )
{
m = k i;
ch = (char)(k i);
cout << ch;
}
}
}
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to print common characters
// of two strings in alphabetical order
import java.io.*;
import java.util.*;
// function to find similar characters
public class simstrings
{
static final int max_char = 26;
static void printcommon(string s1, string s2)
{
// two arrays of length 26 to store occurrence
// of a letters alphabetically for each string
int[] a1 = new int[max_char];
int[] a2 = new int[max_char];
int length1 = s1.length();
int length2 = s2.length();
for (int i = 0 ; i < length1 ; i )
a1[s1.charat(i) - 'a'] = 1;
for (int i = 0 ; i < length2 ; i )
a2[s2.charat(i) - 'a'] = 1;
// if a common index is non-zero, it means
// that the letter corresponding to that
// index is common to both strings
for (int i = 0 ; i < max_char ; i )
{
if (a1[i] != 0 && a2[i] != 0)
{
// find the minimum of the occurrence
// of the character in both strings and print
// the letter that many number of times
for (int j = 0 ; j < math.min(a1[i], a2[i]) ; j )
system.out.print(((char)(i 'a')));
}
}
}
// driver code
public static void main(string[] args) throws ioexception
{
string s1 = "geeksforgeeks", s2 = "practiceforgeeks";
printcommon(s1, s2);
}
}
python 3
# python3 program to print common characters
# of two strings in alphabetical order
# initializing size of array
max_char=26
# function to find similar characters
def printcommon( s1, s2):
# two arrays of length 26 to store occurrence
# of a letters alphabetically for each string
a1 = [0 for i in range(max_char)]
a2 = [0 for i in range(max_char)]
length1 = len(s1)
length2 = len(s2)
for i in range(0,length1):
a1[ord(s1[i]) - ord('a')] = 1
for i in range(0,length2):
a2[ord(s2[i]) - ord('a')] = 1
# if a common index is non-zero, it means
# that the letter corresponding to that
# index is common to both strings
for i in range(0,max_char):
if (a1[i] != 0 and a2[i] != 0):
# find the minimum of the occurrence
# of the character in both strings and print
# the letter that many number of times
for j in range(0,min(a1[i],a2[i])):
ch = chr(ord('a') i)
print (ch, end='')
# driver code
if __name__=="__main__":
s1 = "geeksforgeeks"
s2 = "practiceforgeeks"
printcommon(s1, s2);
# this code is contributed by abhishek sharma
c
// c# program to print common characters
// of two strings in alphabetical order
using system;
// function to find similar characters
public class simstrings
{
static int max_char = 26;
static void printcommon(string s1, string s2)
{
// two arrays of length 26 to store occurrence
// of a letters alphabetically for each string
int[] a1 = new int[max_char];
int[] a2 = new int[max_char];
int length1 = s1.length;
int length2 = s2.length;
for (int i = 0 ; i < length1 ; i )
a1[s1[i] - 'a'] = 1;
for (int i = 0 ; i < length2 ; i )
a2[s2[i] - 'a'] = 1;
// if a common index is non-zero, it means
// that the letter corresponding to that
// index is common to both strings
for (int i = 0 ; i < max_char ; i )
{
if (a1[i] != 0 && a2[i] != 0)
{
// find the minimum of the occurrence
// of the character in both strings and print
// the letter that many number of times
for (int j = 0 ; j < math.min(a1[i], a2[i]) ; j )
console.write(((char)(i 'a')));
}
}
}
// driver code
public static void main()
{
string s1 = "geeksforgeeks", s2 = "practiceforgeeks";
printcommon(s1, s2);
}
}
java 描述语言
输出:
eeefgkors
时间复杂度:如果我们考虑 n =长度(更大的字符串)，那么这个算法以 o(n) 复杂度运行。
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