38. Count and Say

38. Count and Say

The count-and-say sequence is the sequence of integers with the first five terms as following:

1. 12. 113. 214. 12115. 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 term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1Output: "1"

Example 2:

Input: 4Output: "1211"

题意： n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211，依次类推。

class Solution { public String countAndSay(int n) { if(n == 1){ return "1"; } //递归调用，然后对字符串处理 String str = countAndSay(n-1) + "*";//为了str末尾的标记，方便循环读数 char[] c = str.toCharArray(); int count = 1; String s = ""; for(int i = 0; i < c.length - 1;i++){ if(c[i] == c[i+1]){ count++;//计数增加 }else{ s = s + count + c[i];//上面的*标记这里方便统一处理 count = 1;//初始化 } } return s; }}

class Solution { public String countAndSay(int n) { String result="1"; for(int i = 0; i < n-1; i++) { result = helper(result); } return result; } private String helper(String s) { String result = ""; int count = 1; for(int i = 0; i < s.length(); i++) { if (i < s.length()-1 && s.charAt(i) == s.charAt(i+1)) { count++; } else { result = result + count + s.charAt(i); count = 1; } } return result; }}

class Solution { public String countAndSay(int n) { if(n == 1){ return "1"; } //递归调用，然后对字符串处理 String str = countAndSay(n-1) + "*";//为了str末尾的标记，方便循环读数 char[] c = str.toCharArray(); int count = 1; String s = ""; for(int i = 0; i < c.length - 1;i++){ if(c[i] == c[i+1]){ count++;//计数增加 }else{ s = s + count + c[i];//上面的*标记这里方便统一处理 count = 1;//初始化 } } return s; }}

class Solution { public String countAndSay(int n) { if(n == 1) return "1"; String say = countAndSay(n-1); StringBuilder ret = new StringBuilder(); int i = 1, count = 1; for(; i < say.length(); i++,count++) // skip first char if(say.charAt(i-1) != say.charAt(i)){ ret.append(count).append(say.charAt(i - 1)); count = 0; // reset count to zero } return ret.append(count).append(say.charAt(i - 1)).toString(); }}

class Solution { public String countAndSay(int n) { StringBuilder progresString = new StringBuilder(); int counter = 0; int current = 0; String finalString = "1"; if (n == 1) return finalString; for (int i = 1; i < n; i++) { while (counter < finalString.length()) { while ((counter < finalString.length()) && finalString.charAt(current) == finalString.charAt(counter)) { counter++; } progresString = progresString.append(counter - current); progresString = progresString.append(finalString.charAt(current)); current=counter; } finalString = progresString.toString(); counter = 0; current = 0; progresString = new StringBuilder(); } return finalString; }}

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