如何判断一个数字字符串里是否有某个数字

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public class Main {    public static void main(String[] args) {
        String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
        int i = 15;
        int j = 16;
        boolean b1=str.contains(String.valueOf(i));
        boolean b2=str.contains(String.valueOf(j));
        System.out.println(b1);
        System.out.println(b2);
    }
}

public class Main {    public static void main(String[] args) {
        String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
        int i = 15;
        int j = 16;
        boolean b=str.contains(String.valueOf(i))&&!str.contains(String.valueOf(j));
        System.out.println(b);
    }
}

public static void main(String[] args) {
String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
        int i = 15;
        int j = 16;
        boolean b1=str.contains(i+"");
        boolean b2=str.contains(j+"");
        System.out.println(b1);
        System.out.println(b2);
}

楼上的答案都有遗漏，假设这个里面字符串里面是
str="1,12"
我要找是否包含2,，楼上的全部都要找出来，所以嘛，呵呵。这样解决    public static void main(String[] args) {
        String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
        String temp=","+str+",";
        int i = 15;
        int j = 16;
        boolean b1=str.contains(","+i+",");
        boolean b2=str.contains(","+j+",");
        System.out.println(b1);
        System.out.println(b2);
    }

好像我这个不怎么效率，但是是正确的（大家的好像不能判断某些情况吧，比如说i=1的情况）
public static boolean contains(String str, int i){
        String[] gs = str.split(",");
        for(int k=0; k<gs.length; k++){
         if(String.valueOf(i).equalsIgnoreCase(gs[k])){
         return true;
         }
        }
        return false;
}

public class ContainNumber {

public boolean isContainNumber(String str, int num) {
if(str.startsWith(",")) str = ","+str;
if(str.endsWith(",")) str += ",";
/*****************************************************
*  以上2步保证最终的字符串是",1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35,"
*  然后就能用indexOf(",15,") 这样就能唯一了.
*  如果没有开始的逗号的话那么"115," indexOf(",15,")返回的不是-1
*/
if(str.indexOf(""+num)!=-1) return true;
return false;
} public static void main(String[] args) {
ContainNumber number = new ContainNumber();
String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
        System.out.println(number.isContainNumber(str, 15));
        System.out.println(number.isContainNumber(str, 16));
}
}

    public static void main(String[] args) {
        String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
        String temp=","+str+",";
        int i = 15;
        int j = 16;
        boolean b1=temp.contains(","+i+",");
        boolean b2=temp.contains(","+j+",");
        System.out.println(b1);
        System.out.println(b2);
    }

import java.util.ArrayList;public class Main{
    public static void main(String[] args) {
         String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
         int i=15;
         int j=16;
         String strs[]=str.split(",");
         ArrayList<Integer> list=new ArrayList<Integer>();
         for (int k = 0; k < strs.length; k++) {
           list.add(Integer.valueOf(strs[k]));
        }
        boolean b=list.contains(i)&&!list.contains(j);
        System.out.println(b);
    }
}

public class Ss {
public static void main(String args[]){
String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
int i = 15;
int j = 16;

int index=str.indexOf(String.valueOf(i));
int index2=str.indexOf(String.valueOf(j));
System.out.println(index+" "+index2);
}}
index 为－1时则不包含，不为-1时则包含

public static void main(String[] args) {
        String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
        String temp=","+str+",";
        int i = 15;
        int j = 16;
        boolean b1=str.contains(","+i+",");
        boolean b2=str.contains(","+j+",");
        System.out.println(b1);
        System.out.println(b2);
    }

String num = "1,2,3,4,16";
int[] intArr = {1, 6};
for(int i = 0; i < intArr.length; i++){
StringTokenizer st1 = new StringTokenizer(num, ",");
while (st1.hasMoreElements()) {
if (st1.nextElement().equals(new String(new Integer(intArr[i]).toString()))) {
System.out.println("true");
break;
}
}
if(!st1.hasMoreElements()){
System.out.println("false");
}
}

public void compare(){
char[] c = str.toChar();
int length = c.length;
int i = 0;
int a = 15;
int b = 16;
for( int i = 0; i < length; i++　)｛
switch( (int)char[i] ){
case a:
i = 1;
break;
case b:
i = 2;
break;
}
｝
if( i == 1 ){
只有a;
}else if( i == 2 ){
2个都有
}else {
2个都没有
}
}

上面a,b不是赋值15，16，而是先对应char的转成int的值。

indexOf(String.valueOf(i))=-1
如果是数组..就不一样了

上面的做法都对,但是效率太低,你们用的contains这个本身要创建2个对象然后用equals比较，效率肯定低。正确的做法是
public class FindNumber{
public static void main(String[] args)
{
    String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
    int i = 15;
    int j = 16;
    String[] strValue = str.split(",");
    for(int k=0;k<strValue.length;k++)
    {
if(Integer.parseInt(strValue[k]) == i)
{
System.out.println("Y");
return;
}
    }
    System.out.println("N");
}
}

最简单的最有效率的,不用说了,正则表达式.这个题目正则表达式匹配如下:
String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
int i = 15;
int j = 16; Pattern p = Pattern.compile("[15]*");
Matcher m = p.matches(str);
if(m.find()) {
查找到进行相关处理
} else {
查找不到做相关处理。
}

public class Test
{
public static void main(String[] args)
{
String str = "1,2,3,4,5,690,7";
Test test = new Test();
System.out.println(test.check(str,690)); } public boolean check(String str, int arg)
{
String[] arr = str.split(",");
for (String a : arr)
{
if (a.equals(String.valueOf(arg)))
{
return true;
}
}
return false;
}}

public static void main(String[] args){
  String str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
  int i = 15;
  int j = 16;
  for(String s : str.split(",")){
    if(s.equals(i+"")){
      if(!s.equals(j+"")){
        System.out.println("ok");
      }
    }
  }
}

[Quote=引用 26 楼 xiao5122122455 的回复:]
引用 25 楼 lgg201 的回复:
最简单的最有效率的,不用说了,正则表达式. 这个题目正则表达式匹配如下: Java codeString str = "1,2,3,4,5,6,7,8,11,14,15,18,22,24,27,30,33,35";
int i = 15;
int j = 16; Pattern p = Pattern.compile("[15]*");
Matcher m = p.matches(str);
if(m.find()) {
查找到进行相关处理
} else {
查找不到做相关处理。
}
如果str里有个151呢?那不是也匹配?
这种除了把str用split成数组,其它的办法都不怎么合适

这是你前提知道 String 里有 “，”。假如不知道呢或者没有
那怎么使用 split
比如 String str=“abd12ds13ed”;让你判断字符串里是否有数字

if(a.indexOf(b)!=-1)
则表示a包含b

调试易

如何判断一个数字字符串里是否有某个数字

解决方案 »

如何判断一个数字 字符串里是否有某个数字

解决方案 »

如何判断一个数字字符串里是否有某个数字