The count-and-say sequence is the sequence of integers with the first five terms as following:

1.     1
2.     11
3.     21
4.     1211
5.     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 where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: 4
Output: "1211"
Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 
and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".

外观数列。题意是根据规律输出数字。规律如下，

• 首先1，自然表示为1，而say出来就是1个1
• 所以2应该表示为11，而say出来就是2个1
• 所以3应该表示为21，而say出来就是1个2，1个1
• 所以4应该表示为1211，以此类推
• 5 - 111221
• 6 - 312211

这个题没有什么好的思路，就只能按照规则一个个算下一个数是多少，直到第N个数字。首先第一个数字一定是1，所以可以先用一个变量res记录下来，然后之后的数字都依据遍历第一个数字的结果来记录。直接看代码应该可以懂。

时间O(1) - 因为时间复杂度并不完全根据input大小有线性关系

空间O(n)

Java实现

 1 class Solution {
 2     public String countAndSay(int n) {
 3         int i = 1;
 4         String res = "1";
 5         while (i < n) {
 6             int count = 0;
 7             StringBuilder sb = new StringBuilder();
 8             char c = res.charAt(0);
 9             for (int j = 0; j <= res.length(); j++) {
10                 if (j != res.length() && res.charAt(j) == c) {
11                     count++;
12                 } else {
13                     sb.append(count);
14                     sb.append(c);
15                     if (j != res.length()) {
16                         count = 1;
17                         c = res.charAt(j);
18                     }
19                 }
20             }
21             res = sb.toString();
22             i++;
23         }
24         return res;
25     }
26 }

 

JavaScript实现

 1 /**
 2  * @param {number} n
 3  * @return {string}
 4  */
 5 var countAndSay = function (n) {
 6     let i = 1;
 7     let res = '1';
 8     while (i < n) {
 9         let count = 0;
10         let sb = [];
11         let c = res.charAt(0);
12         for (let j = 0; j <= res.length; j++) {
13             if (j != res.length && res.charAt(j) === c) {
14                 count++;
15             } else {
16                 sb.push(count);
17                 sb.push(c);
18                 if (j != res.length) {
19                     count = 1;
20                     c = res.charAt(j);
21                 }
22             }
23         }
24         res = sb.join('');
25         i++;
26     }
27     return res;
28 };

 

LeetCode 题目总结