Count and Say

最新推荐文章于 2024-12-21 10:31:21 发布

flyatcmu

最新推荐文章于 2024-12-21 10:31:21 发布

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分类专栏： Two pointers

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本文介绍了一种生成计数读数序列的方法，通过双指针技术递归地构建序列的每一项。从初始项“1”开始，通过读取并计数相同数字的连续出现次数来生成后续项。

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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".

思路：就是个双指针的题，每次扫描的结果，是下次的输入；

class Solution {
    public String countAndSay(int n) {
        String res = "1";
        for(int i = 1; i < n; i++) {
            StringBuilder sb = new StringBuilder();
            int j = 0;
            while(j < res.length()) {
                int count = 1;
                while(j + 1 < res.length() && res.charAt(j) == res.charAt(j + 1)) {
                    j++;
                    count++;
                }
                sb.append(count);
                sb.append(res.charAt(j));
                j++;
            }
            res = sb.toString();
        }
        return res;
    }
}