Count and Say

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.

 

思路

直接模拟的方法。

 1     string countAndSay(int n) {
 2         // Note: The Solution object is instantiated only once and is reused by each test case.
 3         string s = "1", t;
 4         if(n == 1)
 5             return s;
 6         int i, j;
 7         for(i = 1; i < n; i++){
 8             t = "";
 9             int l = s.length();
10             for(j = 0; j < l; j++){
11                 if(((j+1<l)&&s[j+1]!=s[j]) || (j+1 >= l)){
12                     t += '1';
13                     t += s[j];
14                 }
15                 else{
16                     int num = 1;
17                     while((j+num<l) && (s[j+num]==s[j])){
18                         num++;
19                     }
20                     j+=(num-1);
21                     if(num>999){
22                         t += (char)(num/1000 + '0');
23                         num = num%1000;
24                     }
25                     if(num>99){
26                         t += (char)(num/100 + '0');
27                         num = num%100;
28                     }
29                     if(num>9){
30                         t += (char)(num/10 + '0');
31                         num = num%10;
32                     }
33                     t += (char)(num + '0');  
34                     t += s[j];                  
35                 }
36             }
37             s = t;
38         }
39         return s;
40     }

 第二遍

 1     string countAndSay(int n) {
 2         string result = "1";
 3         if(n == 1)
 4             return result;
 5         string next;
 6         int i,j,k;
 7         for(i = 2; i <= n; i++){
 8             j = 0;
 9             while(j < result.length()){
10                 k = j+1;
11                 while(k < result.length() && result[k] == result[j])
12                     k++;
13                 next += (k-j+'0');
14                 next += result[j];
15                 j = k;
16             }
17             result = next;
18             next = "";
19         }
20         return result;
21     }

 

posted on 2013-10-15 14:35  waruzhi  阅读(200)  评论(0)  编辑  收藏  举报

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