java11-6 String类的其它功能


String类的其他功能:

  替换功能:
    String replace(char old,char new)
    String replace(String old,String new)

  去除字符串两空格
    String trim()

  按字典顺序比较两个字符串
    int compareTo(String str) 区分大小写
    int compareToIgnoreCase(String str) 不区分大小写

 1 public class StringTest3 {
 2 
 3     public static void main(String[] args) {
 4         String s = "HelloWorld";
 5         
 6         // 字符替代:String replace(char old,char new)
 7         String s1 = s.replace('o','k');
 8         System.out.println(s1);//HellkWkrld
 9         
10         // 字符串替代:String replace(String old,String new)
11         String s2 = s.replace("ll", "kk");
12         System.out.println(s2);//HekkoWorld
13         
14         
15         // 去除字符串两空格
16         String s4 = " hello world  ";
17         String s5 = s4.trim();
18         System.out.println("s4:" + s4 + "---");//左右的空格还保留
19         System.out.println("s5:" + s5 + "---");//左右的空格去除了
20 
21         
22         //按字典顺序(ASCII码表)比较两个字符串:
23         // int compareTo(String str)
24         String a1 = "cat";
25         String a2 = "dog";
26         String a3 = "Cat";
27         String a4 = "door";
28         String a5 = "cat";
29         String a6 = "c";
30         
31         int i1 = a1.compareTo(a2);
32         System.out.println(i1);
33         //-1 c在b的前面一位,ASCII码表,a1的c减去a2的b,等于-1
34         
35         System.out.println(a1.compareTo(a3));//32
36         //a1的c在ASCII的数值减去a2的C在ASCII的数值。得到32
37         
38         System.out.println(a2.compareTo(a4));//-8
39         //a2的前2个字母do和a4前2个一样,a2第三个的g减去a4第三个的o,得到-8
40         
41         System.out.println(a1.compareTo(a5));//0
42         //a1和a5的数组元素完全一样
43         
44         System.out.println(a1.compareTo(a6));//2
45         //Ctrl+左键点击comperTo看源码
46     }
47 
48 }
49 
50 compareTo的源码,以及“System.out.println(a1.compareTo(a6));//2”的分析
51 
52  private final char value[];
53   
54     字符串会自动转换为一个字符数组。
55   
56   
57 public int compareTo(String anotherString) {
58         // this - a1 - "cat"
59         // anotherString - a2 - "c"
60         
61         int len1 = value.length; 
62         //this.value.length = a1.toCharArray().length == 3 
63        
64         int len2 = anotherString.value.length;
65         //anotherString.value.length = a2.toCharArray().length == 1
66         
67         int lim = Math.min(len1, len2);
68         // lim = Math.min = 3-1 = 2; lim = 2
69         
70         char v1[] = value;
71         // char v1[] = {'c','a','t'};
72        
73         char v2[] = anotherString.value;
74         // char v2[] = {'c'};
75         
76         int k = 0;
77         while (k < lim) {
78             char c1 = v1[k]; //c1 = 'c','a','t'
79             char c2 = v2[k]; //c2 = 'c'
80             if (c1 != c2) {
81                 return c1 - c2;
82             }
83             k++; //当k=1时,k !< lim ,lim = 1; 所以跳出循环
84         }
85         return len1 - len2; //跳出循环后,到这里,比较两个数组的长度  3-1 = 2
86     }
87     
88     String a1 = "cat";
89     String a6 = "c";
90     System.out.println(a1.compareTo(a6));//2
何事都只需坚持.. 难? 维熟尔。 LZL的自学历程...只需坚持
原文地址:https://www.cnblogs.com/LZL-student/p/5875711.html