LeetCode 38 -- Count and Say

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本文解析了LeetCode第38题Count and Say的问题，详细介绍了递归和迭代两种方法来生成特定序列的第N项，并提供了完整的Java代码实现。

LeetCode 38 -- 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.

解答：本题需要读懂题意，根据上面的count和say的规则，生成一个序列，返回该序列的第N个。程序实现时候，当字符与上一个字符相同时，只计数，不同时，添加上一个字符和上一个字符对应的计数，然后重新计数。第N个序列是根据第N-1个序列产生，因此我们使用递归的方法。

递归方法：

<span style="font-size:14px;">public class Solution {
    public String countAndSay(int n) {
        if(n == 0) return("");
        if(n == 1) return("1");
        
        String s = countAndSay(n - 1);
        char ch = s.charAt(0);
        int count = 1;
        StringBuilder sb = new StringBuilder();
        for(int i = 1; i < s.length(); i++){
            if(s.charAt(i) == ch){
                count ++;
            }    
            else{
                sb.append(count);
                sb.append(ch);
                count = 1;
                ch = s.charAt(i);
            }
        }
        sb.append(count);
        sb.append(ch);
        
        return sb.toString();
    }
}</span>

迭代方法：

<span style="font-size:14px;">public class Solution {
    public String countAndSay(int n) {
        if(n ==0)   return "";
        String s ="1";
        
        for(int j = 1; j < n; j++){
            StringBuilder sb = new StringBuilder();
            char last = s.charAt(0);
            int count = 1;
            for(int i = 1; i < s.length(); i++){
                if(s.charAt(i) ==  last)    count++;
                else{
                    sb.append(count);
                    sb.append(last);
                    last = s.charAt(i);
                    count = 1;
                }
            }
            sb.append(count);
            sb.append(last);
            s = sb.toString();
        }
        return s;
    }
}</span>