How to implement the appearance sequence of Go Java algorithm
Appearance sequence
Given a positive integer n, output the nth item of the appearance sequence.
"Appearance sequence" is a sequence of integers, starting from the number 1, and each item in the sequence is a description of the previous item.
You can think of it as a sequence of numeric strings defined by a recursive formula:
countAndSay(1) = "1"
countAndSay(n) is a description of countAndSay(n-1), which is then converted into another numeric string.
The first five items are as follows:
1, 1 —— The first item is the number 1
2, 11 - Describe the previous item, this number is 1, that is, "a 1", recorded as "11"
3, 21 - Describe the previous item item, this number is 11, which is "two 1s", recorded as "21"
4, 1211 - describe the previous item, this number is 21, which is "one 2" One 1", recorded as "1211"
- ##5, 111221 - Describe the previous item, this number is 1211, that is, "one 1, one 2, two 1", recorded as "111221"
countAndSay(1) = "1";countAndSay(n) is Description of countAndSay(n-1) and then converted to another numeric string.We define the string S_{i} to represent countAndSay(i). If we require S_{n}, we need to find S_{n-1} first, and then follow the above description Method generation, that is, scanning the maximum number of consecutive identical characters in the string S_{n-1} from left to right, then converting the statistical number of characters into a numerical string and concatenating the corresponding characters.
class Solution { public String countAndSay(int n) { String str = "1"; for (int i = 2; i <= n; ++i) { StringBuilder sb = new StringBuilder(); int start = 0; int pos = 0; while (pos < str.length()) { while (pos < str.length() && str.charAt(pos) == str.charAt(start)) { pos++; } sb.append(Integer.toString(pos - start)).append(str.charAt(start)); start = pos; } str = sb.toString(); } return str; } }
func countAndSay(n int) string { if n == 1 { return "1" } s := countAndSay(n - 1) i, res := 0, "" length := len(s) for j := 0; j < length; j++ { if s[j] != s[i] { res += strconv.Itoa(j-i) + string(s[i]) i = j } } res += strconv.Itoa(length-i) + string(s[i]) return res }
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