《LeetCode之每日一题》:178.外观数列

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外观数列


题目链接: 外观数列

有关题目

给定一个正整数 n ,输出外观数列的第 n 项。

「外观数列」是一个整数序列,从数字 1 开始,序列中的每一项都是对前一项的描述。

你可以将其视作是由递归公式定义的数字字符串序列:

countAndSay(1) = "1"
countAndSay(n) 是对 countAndSay(n-1) 的描述,然后转换成另一个数字字符串。

前五项如下:
1.     1
2.     11
3.     21
4.     1211
5.     111221
第一项是数字 1 
描述前一项,这个数是 1 即 “ 一 个 1 ”,记作 "11"
描述前一项,这个数是 11 即 “ 二 个 1 ” ,记作 "21"
描述前一项,这个数是 21 即 “ 一 个 2 + 一 个 1 ” ,记作 "1211"
描述前一项,这个数是 1211 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ,记作 "111221"
要 描述 一个数字字符串,首先要将字符串分割为 最小 数量的组,每个组都由连续的最多 相同字符 组成。
然后对于每个组,先描述字符的数量,然后描述字符,形成一个描述组。
要将描述转换为数字字符串,先将每组中的字符数量用数字替换,再将所有描述组连接起来。

示例 1:

输入:n = 1
输出:"1"
解释:这是一个基本样例。
示例 2:

输入:n = 4
输出:"1211"
解释:
countAndSay(1) = "1"
countAndSay(2) ="1" = 一 个 1 = "11"
countAndSay(3) ="11" = 二 个 1 = "21"
countAndSay(4) ="21" = 一 个 2 + 一 个 1 = "12" + "11" = "1211"
提示:

1 <= n <= 30

题解

法一:模拟
代码一:

class Solution {
public:
    string countAndSay(int n) {
        string ans = "1";

        while(--n)
        {
            string temp = "";
            int cnt = 1;
            for (int i = 0; i < ans.size(); i++)
            {
                if (i + 1 < ans.size() && ans[i] == ans[i + 1])
                    cnt++;
                else 
                {
                    temp += to_string(cnt) + ans[i];
                    cnt = 1;
                }
            }
            ans = temp;
        }

        return ans;
    }
};

代码二:
参考官方题解

class Solution {
public:
    string countAndSay(int n) {
        string prev = "1";

        for (int i = 2; i <= n; i++)
        {
            string curr = "";
            int pos = 0, start = 0;
            while(pos < prev.size())
            {
                while(pos < prev.size() && prev[pos] == prev[start])
                    pos++;
                
                curr += to_string(pos - start) + prev[start];
                start = pos;
            }
            prev = curr;
        }

        return prev;
    }
};

代码三:
参考官方题解评论区下shira_yuki
Tips

std::adjacent_find(First, Last, p)
作用
	在P的条件下,[First, Last)范围内寻找首个满足p的元素
	
std::exchange(obj, new_value)
以 new_value 替换 obj 的值,并返回 obj 的旧值。
class Solution {
public:
    string countAndSay(int n) {
        string ans = "1", temp = "";

        for (; --n; ans = exchange(temp, ""))
            for (auto i = begin(ans), e = end(ans); i != e; )
            {
                auto j = adjacent_find(i, e, not_equal_to{});
                auto t = exchange(i, (j == e ? j : next(j)));
                temp += to_string(i - t) + *t;
            }

            return ans;
    }
};

时间复杂度:O(n * m), m 为生成的字符串中的最大长度
空间复杂度:O(m),m 为生成的字符串中的最大长度

法二:打表
参考官方题解

class Solution {
public:
    string countAndSay(int n) {
        vector<string> arr = 
        {
            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        };
        return arr[n];        
    }
};

以上是关于《LeetCode之每日一题》:178.外观数列的主要内容,如果未能解决你的问题,请参考以下文章

每日一题篇 — leetcode38号题之外观数列

leetcode 每日一题 38. 外观数列

《LeetCode之每日一题》:120.等差数列划分

LeetCode 剑指 Offer II 069. 山峰数组的顶部(三分) / 38. 外观数列 / 282. 给表达式添加运算符

[每日一题2020.06.14]leetcode #70 爬楼梯 斐波那契数列 记忆化搜索 递推通项公式

《LeetCode之每日一题》:280.矩阵置零