Count and Say 数数并打印 @LeetCode

题目：

数数并打印

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    因为前面一行有1个1
2 2 1     因为前面一行有2个1
3 1 2 1 1  因为前面一行有1个2和1个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"

思路：

遍历，每次对前面的string进行分析，输出

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retret.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();}}

public class Solution {    public String countAndSay(int n) {        String s = "1";                for(int i=1; i<n; i++) {            int cnt = 1;            StringBuilder tmp = new StringBuilder();            for(int j=1; j<s.length(); j++) {                if(s.charAt(j) == s.charAt(j-1)) {                    cnt++;                } else {                    tmp.append(cnt).append(s.charAt(j-1));                    cnt = 1;                }            }            tmp.append(cnt).append(s.charAt(s.length()-1));            s = tmp.toString();        }                return s;    }}

注意到这题如果用 tmp.append(cnt+s.charAt(j-1)); 则会超时