我想匹配兩個字符串並獲得按順序匹配多少個字符。 像:在兩個字符串的序列中匹配的字符數
String one="ABCDEFGHIJK";
String two="ZANDEFGHOPQ";
就像你可以看到,它必須返回序列5個字符相匹配,因爲DEFGH兩個序列存在。
感謝
我想匹配兩個字符串並獲得按順序匹配多少個字符。 像:在兩個字符串的序列中匹配的字符數
String one="ABCDEFGHIJK";
String two="ZANDEFGHOPQ";
就像你可以看到,它必須返回序列5個字符相匹配,因爲DEFGH兩個序列存在。
感謝
試試這個
String one="ABCDEFGHIJK";
String two="ZANDEFGHOPQ";
int cnt=0;
for (int i=0;i<one.length();i++){
for (int j=0;j<two.length();j++){
if(one.charAt(i) == two.charAt(j)){
cnt++;
}
}
}
Toast.makeText(this, "count"+cnt, Toast.LENGTH_SHORT).show();
謝謝@Nilesh讓我檢查這一個 –
最歡迎@AmitSharma –
也許這樣的事情一定會做到。
String one="ABCDEFGHIJK";
String two="ZANDEFGHOPQ";
int counter = 0;
// Iterate over the string character by character - stop when reaching the
// end of the shortest string
for(int i=0; i<one.length() && i<two.length(); i++) {
// Compare the strings character at the current position/index
if(one.charAt(i) == two.charAt(i)) {
// The characters matched so increment the counter
counter++;
}
}
import java.io.*;
import java.util.*;
class happy {
public static void main(String args[])
{
String one="ABCDEFGHIJK";
String two="ZANDEFGHOPQ";
int a=0,c=0;
if(one.length()<two.length())
a=one.length();
else
a=two.length();
for(int i=0;i<a;i++)
{
if(one.charAt(i)==(two.charAt(i)))
c++;
}
System.out.println(c);
}
}
public class Test {
public static void main(String[] args) {
String one="ABCDEFGHIJK";
String two="ZANDEFGHOPQ";
int size1 = one.length();
int size2 = two.length();
int i = 0;
int j = (size1 >= size2 ? size2 : size1);
char[] oneTab = one.toCharArray();
char[] twoTab = two.toCharArray();
int k = 0;
for(i = 0; i< j; i++){
if((String.valueOf(oneTab[i])).equals(String.valueOf(twoTab[i]))){
k = k+1;
};
};
System.out.println(k);
}
}
的[簡單的方式將字符串中的字符數出現(可能的複製https://stackoverflow.com/questions/6100712/simple-way-to- count-character-occurrences-a-string) –
@PratikTank我檢查了這一個,但問題是它只計算單個章程哪個用戶必須知道,它不符合我的需要 –
可能重複的[查找字符出現次數在一個Java字符串](https://stackoverflow.com/questions/3763970/find-occurrences-of-c haracters-in-a-java-string) –