DAY30：演算法題 Count and Say（轉）

題目：

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.

• 題意：給定一個整數字元串，下一個是對上一個的描述，比如」11122211「，下一個是3個1,3個2,2個1，是」313221「，給定n，求出第n個整數字元串
• 關鍵在於找到當前和前一個的關係，比較一個整數字元串的元素，緊鄰是否相同，用變數num統計連續相同的個數，當不想同時候，stringBuffer.append(num).append(元素)，num= 1；相同時，num++；

代碼：

public class Solution { public String countAndSay(int n) { String str = "1"; for(int i = 1;i < n;i++){ int num = 1; StringBuffer sb = new StringBuffer(); for(int j = 1;j < str.length();j++){ if(str.charAt(j) == str.charAt(j-1)){ num++; }else{ sb.append(num).append(str.charAt(j-1)); num = 1; } } sb.append(num).append(str.charAt(str.length()-1)); str = sb.toString(); } return str; }}

推薦閱讀：