Count and Say

Count and Say
- 题目描述
- 解法

Count and Say

题目描述

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.

解法

代码如下:

public static String countAndSay(int n) {
    String res = "1";
    for(int i = 1; i < n; i++) {
        int low = 0, high = 1;
        String tmp = "";
        while(high <= res.length()) {
            while(high < res.length() && res.charAt(high) == res.charAt(high-1)) 
                high++;
            int count = high - low;
            tmp = tmp + count + res.charAt(low);
            low = high++;
        }
        res = tmp;
    }
    return res;
}

思考过程:

按照规律模拟生成.

时空复杂度: 时间复杂度是O(n^2), 空间复杂度是O(n^2)