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
