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字符串和整形数组的相互转化(JAVA程序)

2014-12-22 15:52 423 查看
package te;
public class StringConvert {
static int[] a = {0,1,1,0,1,1,0,2};
static String s = "0011223344";
public static void main(String[] args) {
StringConvert sc = new StringConvert();
System.out.println(sc.intArray2Str(a));
int[] b = sc.str2IntArray(s);
for(int i=0; i<b.length; i++) {
System.out.print(b[i]);
}
}

String intArray2Str(int[] a) {

int len = a.length;
String str="";
for(int i=0;i<len; i++) {
str+=String.valueOf(a[i]);
}
return str;
}

int[] str2IntArray(String str) {
int len = str.length();
int[] a = new int[len];
char[] c = str.toCharArray();
for(int i=0; i<len; i++) {
a[i] = c[i]-'0';
}
return a;
}

}


结果

01101102
0011223344


package te;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class StringConvert {
static int[] a = {0,1,1,0,1,1,0,2};
static String s = "0011223340004";
public static void main(String[] args) {
StringConvert sc = new StringConvert();
System.out.println(sc.intArray2Str(a));
int[] b = sc.str2IntArray(s);
for(int i=0; i<b.length; i++) {
System.out.print(b[i]);
}
System.out.print('\n');
sc.indexsof(s);
}

String intArray2Str(int[] a) {

int len = a.length;
String str="";
for(int i=0;i<len; i++) {
str+=String.valueOf(a[i]);
}
return str;
}

int[] str2IntArray(String str) {
int len = str.length();
int[] a = new int[len];
char[] c = str.toCharArray();
for(int i=0; i<len; i++) {
a[i] = c[i]-'0';
}
return a;
}
int[] indexsof(String s) {

Pattern p = Pattern.compile("[0]");
Matcher m = p.matcher(s);
while(m.find()) {
System.out.println(m.start());
}

return null;

}
}


从一个字符串中找到符合要求字符串的所有位置
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