38. 外观数列

题目描述

「外观数列」是一个数位字符串序列，由递归公式定义：

countAndSay(1) = "1"
countAndSay(n) 是 countAndSay(n-1) 的行程长度编码。

行程长度编码（RLE）是一种字符串压缩方法，其工作原理是通过将连续相同字符（重复两次或更多次）替换为字符重复次数（运行长度）和字符的串联。例如，要压缩字符串 "3322251" ，我们将 "33" 用 "23" 替换，将 "222" 用 "32" 替换，将 "5" 用 "15" 替换并将 "1" 用 "11" 替换。因此压缩后字符串变为 "23321511"。

给定一个整数 n ，返回 外观数列 的第 n 个元素。

示例 1：

输入：n = 4

输出："1211"

解释：

countAndSay(1) = "1"

countAndSay(2) = "1" 的行程长度编码 = "11"

countAndSay(3) = "11" 的行程长度编码 = "21"

countAndSay(4) = "21" 的行程长度编码 = "1211"

示例 2：

输入：n = 1

输出："1"

解释：

这是基本情况。

提示：

1 <= n <= 30

进阶：你能迭代解决该问题吗？

解法

方法一

Python3JavaC++GoTypeScriptRustJavaScriptC#PHP

class Solution:
    def countAndSay(self, n: int) -> str:
        s = '1'
        for _ in range(n - 1):
            i = 0
            t = []
            while i < len(s):
                j = i
                while j < len(s) and s[j] == s[i]:
                    j += 1
                t.append(str(j - i))
                t.append(str(s[i]))
                i = j
            s = ''.join(t)
        return s

class Solution {
    public String countAndSay(int n) {
        String s = "1";
        while (--n > 0) {
            StringBuilder t = new StringBuilder();
            for (int i = 0; i < s.length();) {
                int j = i;
                while (j < s.length() && s.charAt(j) == s.charAt(i)) {
                    ++j;
                }
                t.append((j - i) + "");
                t.append(s.charAt(i));
                i = j;
            }
            s = t.toString();
        }
        return s;
    }
}

class Solution {
public:
    string countAndSay(int n) {
        string s = "1";
        while (--n) {
            string t = "";
            for (int i = 0; i < s.size();) {
                int j = i;
                while (j < s.size() && s[j] == s[i]) ++j;
                t += to_string(j - i);
                t += s[i];
                i = j;
            }
            s = t;
        }
        return s;
    }
};

func countAndSay(n int) string {
    s := "1"
    for k := 0; k < n-1; k++ {
        t := &strings.Builder{}
        i := 0
        for i < len(s) {
            j := i
            for j < len(s) && s[j] == s[i] {
                j++
            }
            t.WriteString(strconv.Itoa(j - i))
            t.WriteByte(s[i])
            i = j
        }
        s = t.String()
    }
    return s
}

function countAndSay(n: number): string {
    let s = '1';
    for (let i = 1; i < n; i++) {
        let t = '';
        let cur = s[0];
        let count = 1;
        for (let j = 1; j < s.length; j++) {
            if (s[j] !== cur) {
                t += `${count}${cur}`;
                cur = s[j];
                count = 0;
            }
            count++;
        }
        t += `${count}${cur}`;
        s = t;
    }
    return s;
}

use std::iter::once;

impl Solution {
    pub fn count_and_say(n: i32) -> String {
        (1..n)
            .fold(vec![1], |curr, _| {
                let mut next = vec![];
                let mut slow = 0;
                for fast in 0..=curr.len() {
                    if fast == curr.len() || curr[slow] != curr[fast] {
                        next.extend(once((fast - slow) as u8).chain(once(curr[slow])));
                        slow = fast;
                    }
                }
                next
            })
            .into_iter()
            .map(|digit| (digit + b'0') as char)
            .collect()
    }
}

const countAndSay = function (n) {
    let s = '1';

    for (let i = 2; i <= n; i++) {
        let count = 1,
            str = '',
            len = s.length;

        for (let j = 0; j < len; j++) {
            if (j < len - 1 && s[j] === s[j + 1]) {
                count++;
            } else {
                str += `${count}${s[j]}`;
                count = 1;
            }
        }
        s = str;
    }
    return s;
};

using System.Text;
public class Solution {
    public string CountAndSay(int n) {
        var s = "1";
        while (n > 1)
        {
            var sb = new StringBuilder();
            var lastChar = '1';
            var count = 0;
            foreach (var ch in s)
            {
                if (count > 0 && lastChar == ch)
                {
                    ++count;
                }
                else
                {
                    if (count > 0)
                    {
                        sb.Append(count);
                        sb.Append(lastChar);
                    }
                    lastChar = ch;
                    count = 1;
                }
            }
            if (count > 0)
            {
                sb.Append(count);
                sb.Append(lastChar);
            }
            s = sb.ToString();
            --n;
        }
        return s;
    }
}

class Solution {
    /**
     * @param integer $n
     * @return string
     */

    function countAndSay($n) {
        if ($n <= 0) {
            return '';
        }

        $result = '1';
        for ($i = 2; $i <= $n; $i++) {
            $count = 1;
            $say = '';
            for ($j = 1; $j < strlen($result); $j++) {
                if ($result[$j] == $result[$j - 1]) {
                    $count++;
                } else {
                    $say .= $count . $result[$j - 1];
                    $count = 1;
                }
            }
            $say .= $count . $result[strlen($result) - 1];
            $result = $say;
        }
        return $result;
    }
}

38. 外观数列

题目描述

解法

方法一

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