Count and Say @LeetCode

计数并描述序列生成

最新推荐文章于 2025-11-30 19:25:58 发布

最新推荐文章于 2025-11-30 19:25:58 发布 · 84 阅读

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#数据结构与算法

本文介绍了一种特殊的数列生成方法——计数并描述序列，并提供了一个Java实现示例。该序列从数字1开始，后续每一项都是对其前一项的描述。例如，1被描述为“一个1”，即11；11被描述为“两个1”，即21。文章通过迭代方法生成指定位置的序列。

题目比较不直观，这里的描述比较好一些 http://www.careercup.com/question?id=4425679

"Count and Say problem" Write a code to do following:
n String to print
0 1
1 1 1
2 2 1
3 1 2 1 1
...
Base case: n = 0 print "1"
for n = 1, look at previous string and write number of times a digit is seen and the digit itself. In this case, digit 1 is seen 1 time in a row... so print "1 1"
for n = 2, digit 1 is seen two times in a row, so print "2 1"
for n = 3, digit 2 is seen 1 time and then digit 1 is seen 1 so print "1 2 1 1"
for n = 4 you will print "1 1 1 2 2 1"

Consider the numbers as integers for simplicity. e.g. if previous string is "10 1" then the next will be "1 10 1 1" and the next one will be "1 1 1 10 2 1"

package Level2;

/**
 * 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.
 *
 */
public class S38 {

	public static void main(String[] args) {
		System.out.println(countAndSay(6));
	}
	
	public static String countAndSay(int n) {
		
		if(n == 1){
			return "1";
		}
		
		String s = "1";
		StringBuffer ret = new StringBuffer();
		int cnt = 0;
		int round = 0;			// round是迭代多少次
		int i;
		while(++round < n){
			cnt = 1;
			ret.setLength(0);
			for(i=1; i<s.length(); i++){
				if(s.charAt(i) == s.charAt(i-1)){		// 重复的值，继续计数
					cnt++;
				}else{			// 有新的值出现，记录到ret
					ret.append(cnt).append(s.charAt(i-1));
					cnt = 1;		// 重置cnt
				}
			}
			ret.append(cnt).append(s.charAt(i-1));
			s = ret.toString();	// 更新s
		}
		return ret.toString();
	}

}