字符串和整形数组的相互转化（JAVA程序）

 1 package te;
 2 public class StringConvert {
 3     static int[] a = {0,1,1,0,1,1,0,2};
 4     static String s = "0011223344";
 5     public static void main(String[] args) {
 6         StringConvert sc = new StringConvert();
 7         System.out.println(sc.intArray2Str(a));
 8         int[] b = sc.str2IntArray(s);
 9         for(int i=0; i<b.length; i++) {
10             System.out.print(b[i]);
11         }
12     }
13
14     String intArray2Str(int[] a) {
15
16         int len = a.length;
17         String str="";
18         for(int i=0;i<len; i++) {
19             str+=String.valueOf(a[i]);
20         }
21         return str;
22     }
23
24     int[] str2IntArray(String str) {
25         int len = str.length();
26         int[] a = new int[len];
27         char[] c = str.toCharArray();
28         for(int i=0; i<len; i++) {
29             a[i] = c[i]-‘0‘;
30         }
31         return a;
32     }
33
34 }

结果

01101102
0011223344

 1 package te;
 2
 3 import java.util.regex.Matcher;
 4 import java.util.regex.Pattern;
 5
 6 public class StringConvert {
 7     static int[] a = {0,1,1,0,1,1,0,2};
 8     static String s = "0011223340004";
 9     public static void main(String[] args) {
10         StringConvert sc = new StringConvert();
11         System.out.println(sc.intArray2Str(a));
12         int[] b = sc.str2IntArray(s);
13         for(int i=0; i<b.length; i++) {
14             System.out.print(b[i]);
15         }
16         System.out.print(‘\n‘);
17         sc.indexsof(s);
18     }
19
20     String intArray2Str(int[] a) {
21
22         int len = a.length;
23         String str="";
24         for(int i=0;i<len; i++) {
25             str+=String.valueOf(a[i]);
26         }
27         return str;
28     }
29
30     int[] str2IntArray(String str) {
31         int len = str.length();
32         int[] a = new int[len];
33         char[] c = str.toCharArray();
34         for(int i=0; i<len; i++) {
35             a[i] = c[i]-‘0‘;
36         }
37         return a;
38     }
39     int[] indexsof(String s) {
40
41          Pattern p = Pattern.compile("[0]");
42          Matcher m = p.matcher(s);
43          while(m.find()) {
44              System.out.println(m.start());
45          }
46
47         return null;
48
49     }
50 }

从一个字符串中找到符合要求字符串的所有位置