【每日力扣19】外观数列
作者：互联网
一、题目[LeetCode-38]
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。
例如，数字字符串 "3322251" 的描述如下图：
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) = 读 "1" = 一 个 1 = "11"
countAndSay(3) = 读 "11" = 二 个 1 = "21"
countAndSay(4) = 读 "21" = 一 个 2 + 一 个 1 = "12" + "11" = "1211"
提示：
1 <= n <= 30
二、思路
双指针法
本题与《Day1 剔除数组中重复元素》思想类似，可以使用快慢指针，块指针向前遍历数组，当发现前方有不一致元素时慢指针更新，只不过更新方式不一样，《Day1 剔除数组中重复元素》是直接j++；而本题便是j=i+1，因为本题慢指针用于保存相同区域的首元素索引，从而便于后面表示出“这块区域有多少个数”。
class Solution {
public:
string countAndSay(int n) {
if (n == 1)
return "1";//n==1时的基本样例（递归基）
if (n == 2)
return"11";//n==2时的基本样例（递归基）
string s = countAndSay(n - 1);//递归调用前一个结果
int size = s.size();
string m;//这是将要返回的字符串
for (int i = 0, j = 0; i < size - 1; i++)//使用双指针i和j，i向前遍历，j用于保存i所在的相同字符的区域的第一个索引
{
if (s[i] != s[i + 1])//一旦发现连续相同字符的区域结束，下一位开始是新的字符时，
{
m.push_back('0'+i + 1 - j);//存入这片区域的长度
j = i + 1;//并更新慢指针j
m.push_back(s[i]);//存入这片区域的数值
if (i == size - 2)//如果最后一片区域长度恰好为1（该for循环无法遍历到最后一位元素），
{
m.push_back('1');//这里专门存储这片区域
m.push_back(s[i+1]);
}
}
else
if (i == size - 2)//如果最后一片区域长度大于1，
{
m.push_back('0' + size - j);//对其进行存入
m.push_back(s[i]);
}
}
return m;
}
};
三、官方解法（力扣（LeetCode））
方法一：遍历生成（与笔者思路类似，但是代码更简洁）
通过阅读题目，我们可以了解到所谓的「外观数列」，其实本质上只是依次统计字符串中连续相同字符的个数。
例如字符串 1112234445666 我们依次统计连续相同字符的个数为: 3 个连续的字符 1, 2 个连续的 1，1 个连续的字符 3，3 个连续的字符 4，1 个连续的字符 5，3 个连续的字符 6。我们对其进行整理为 (3)1(2)2(1)3(3)4(1)5(3)6，我们将括号内的数字也转换为字符串为 312213341536。
题目中给定的递归公式定义的数字字符串序列如下：
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。
我们定义字符串 Si 表示 countAndSay(i)，我们如果要求得 Sn，则我们需先求出Sn−1，然后按照上述描述的方法生成，即从左到右依次扫描字符串 S
n−1中连续相同的字符的最大数目，然后将字符的统计数目转化为数字字符串再连接上对应的字符。我们已知 S1 为 "1"，我们按照上述方法依次生成即可。
class Solution {
public:
string countAndSay(int n) {
string prev = "1";
for (int i = 2; i <= n; ++i) {
string curr = "";
int start = 0;
int pos = 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;
}
};
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/count-and-say/solution/wai-guan-shu-lie-by-leetcode-solution-9rt8/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。复杂度分析
- 时间复杂度：O(N×M)，其中 N 为给定的正整数，M 为生成的字符串中的最大长度。
- 空间复杂度：O(M)。其中M 为生成的字符串中的最大长度。
方法二：枚举查表（不推荐）
题目中给定的数量限制为 1≤n≤30，由于数量很小，我们只需要依次生成前 30 个 ountAndSay 字符串并将其存储为静态数组，每次直接查询数组中的第 n 个元素即可。
class Solution {
public:
string countAndSay(int n) {
vector<string> arr =
{
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};
return arr[n];
}
};
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/count-and-say/solution/wai-guan-shu-lie-by-leetcode-solution-9rt8/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。复杂度分析
- 时间复杂度：O(1)。
- 空间复杂度：O(C×M)。其中 C 是 N 是上界，在本题中 C=30，M 为生成的字符串中的最大长度。
四、学习心得
双指针的应用
本题的关键在于看懂题目的意思，然后去结合题意使用双指针法。
标签：11,字符,数列,19,力扣,countAndSay,字符串,指针,描述 来源： https://blog.csdn.net/micah_yaokunkun/article/details/122355541