38. 外观数列 @Python @Go

原题

https://leetcode.cn/problems/count-and-say/description/

思路

字典 + 循环

复杂度

时间：O(n)
空间：O(n)

Python代码

class Solution:
    def countAndSay(self, n: int) -> str:
        d = {1: '1'}
        if n == 1:
            return d[1]
        for i in range(2, n + 1):
            s = d[i-1]
            ans = ''
            count = 0 
            char = ''
            for j in range(0, len(s)):
                if j == 0:
                    char = s[j]
                    count += 1 
                else:
                    if s[j] == s[j - 1]:
                        count += 1 
                    else:
                        ans += str(count) + char 
                        count = 1 
                        char = s[j]
            ans += str(count) + char 
            d[i] = ans 
        return d[n]
        

Go代码

func countAndSay(n int) string {
	d := map[int]string{1: "1"}
	if n == 1 {
		return d[1]
	}
	for i := 2; i <= n; i++ {
		// 之前的字符串
		s := d[i-1]
		ans := ""
		count := 0
		var char byte
		for j := 0; j < len(s); j++ {
			if j == 0 {
				char = s[j]
				count++
			} else {
				if s[j] == s[j-1] {
					count++
				} else {
					// 将已压缩的字符加入结果
					ans += string('0'+count) + string(char)
					count = 1
					char = s[j]
				}
			}
		}
		// 将已压缩的字符加入结果
		ans += string('0'+count) + string(char)
		d[i] = ans
	}
	return d[n]
}