【Problem】Count and Say

问题分析

　　题目理解：给定一个整数，如果相邻的几个数位数字是相同的，则输出重复出现的次数加上该数字本身；继续向后查找直到所有的数位处理完。

　　按照上述思路，则：

input	output
1	11
11	21
21	1211

　　但实际运行时提示出错，在输入为1的时候输出结果为11，而实际的应该是1。接着发现题目意思理解错了，如下图所示，后面的输出结果是依赖与前面的输出结果的。比如第一个输出为1；第二个输出是对1进行上述运算，即11；同理，第三个是对11进行运算，即21；接着依次是1211、111221、312211、13112221、1113213211……

源代码

 public class Solution {
     public String countAndSay(int n) {

         if ( n == 1 ) {
             return "1";
         } else {

             char current;    // the current char
             int count;       // the count of the current char
             String result = new String("1");  // the result
             int length = result.length();     // the length of the result string
             StringBuilder strBuilder = new StringBuilder(); // store the current result

             int i = 0;
             while ( n > 1 ) {
                 for ( i = 0; i < length; i++ ) {
                     current = result.charAt(i);
                     count = 1;
                     // while the next char is the same as the current char
                     while ( (i+1 < length) &&
                             (result.charAt(i+1) ==  current) ) {
                         count++;
                         i++;
                     }
                     // add the char and its count to the current result
                     strBuilder.append(count).append(current);
                 }
                 // update the result and its length, and clear the content of strBuilder for the next loop
                 result = strBuilder.toString();
                 length = result.length();
                 strBuilder = new StringBuilder();
                 n--;
             }

             return result;
         }
     }
 }