Java Learning String Practice

1, given a string array. Sort from small to large in dictionary order.

Ideas
* 1, sorted by array. You can use the selection, bubbling is OK.
* 2,for nesting and comparison and transposition.
* 3, Question: The front row is an integer, compared with the comparison operator, but is now a string object.
* How do string objects compare? Cool, the object provides functionality for string object comparisons.

1  Public Static voidsort (string[] strings) {2String temp =NULL;3          for(inti=0;i<strings.length;i++){4              for(intj = i+1;j<=strings.length-1;j++){5                 if(Strings[i].compareto (strings[j]) >0){6temp =Strings[i];7strings[i]=Strings[j];8strings[j]=temp;9                 }Ten             } One         } A}

2. The number of occurrences of a substring in the entire string

Ideas
* 1, the substring to be found is present if there is a location to get it to appear. This can be done using indexof.
* 2, if found, then record the occurrence of the position and continue to find the substring in the remaining string,
* While the starting bit of the remaining string is the length of the position + substring that appears.

* 3, and so on, through the loop to complete the lookup, if not found is-1, and each find with a counter record.

1  Public Static intsum (String string,string string1) {2         intLength =string1.length ();3         intCount = 0;4         intindex =String.IndexOf (string1);5          while(Index!=-1){6count++;7index = String.IndexOf (string1,index+length);8         }9         returncount;Ten}

3, two the largest of the same substring in a string

Ideas
* 1, since the maximum substring is obtained, first look at whether the short string is in the long string.
* If present, the short string is the maximum substring.
* 2, if not, then the short sub-string to the length of the way to the substring, to long string to determine whether there is.
* If it exists, it will not be found.

1  Public Staticstring getmaxsubstring (string s1, string s2) {2         3String max =NULL, Min =NULL;4Max = (S1.length () >s2.length ())?s1:s2;5         6min = max.equals (S1)?s2:s1;7         8System.out.println ("max=" +max);9System.out.println ("min=" +min);Ten          One          A          -          for(inti = 0; I < min.length (); i++) { -              the              for(intA = 0,b = Min.length ()-I; b! = Min.length () +1; a++,b++){ -                  -String Sub =min.substring (A, b); - //System.out.println (sub); +                 if(Max.contains (sub)) -                     returnSub; +             } A         } at          -         return NULL; -}

4, simulate a trim function consistent method. Remove whitespace at both ends of a string
Ideas
* 1, define two variables.
* A variable is used as a starting point to determine the angle of a string space. Constant + +.
* A variable as a corner mark that determines the string space from the end. Constantly--。
* 2, to determine that is not a space, the string between the kinsoku can be.
*/

1  Public Staticstring Mytrim (string s) {2 3         intStart = 0, end = S.length ()-1;4 5          while(Start <= end && S.charat (start) = = ") {6start++;7         }8          while(Start <= end && S.charat (end) = = ") {9end--;Ten         } One         returnS.substring (Start, end + 1); A}

Java Learning String Practice