leetcode-38-外观数列
[博客链接]
[题目链接]
[github地址]
[题目描述]
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
-
- 11
-
- 21
-
- 1211
-
- 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
思路一:打表
- 因为数据集很小只有30个所以我们可以很轻易的打表解决
class Solution {
public String countAndSay(int n) {
String[] arr = {
"","1","11","21","1211","111221","312211","13112221","1113213211","31131211131221","13211311123113112211","11131221133112132113212221","3113112221232112111312211312113211","1321132132111213122112311311222113111221131221","11131221131211131231121113112221121321132132211331222113112211","311311222113111231131112132112311321322112111312211312111322212311322113212221","132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211","11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221","31131122211311123113321112131221123113112211121312211213211321322112311311222113311213212322211211131221131211132221232112111312111213111213211231131122212322211331222113112211","1321132132211331121321231231121113112221121321132122311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112111331121113122112132113213211121332212311322113212221","11131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113121113123112112322111213211322211312113211","311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311122122111312211213211312111322211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221131112311311121321122112132231121113122113322113111221131221","132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113213221133122112231131122211211131221131112311332211211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211","111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122211331121321232221121113122113121113222123112221221321132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322212321121113122123211231131122113221123113221113122112132113213211121332212311322113212221","3113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213213211221113122113121113222112132113213221232112111312111213322112132113213221133112132123123112111311222112132113311213211221121332211231131122211311123113321112131221123113112221132231131122211211131221131112311332211213211321223112111311222112132113212221132221222112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211132221121311121312211213211312111322211213211321322113311213212322211231131122211311123113321112131221123113112211121312211213211321222113222112132113223113112221121113122113121113123112112322111213211322211312113211","1321132132211331121321231231121113112221121321133112132112312321123113112221121113122113111231133221121321132122311211131122211213211321322112312321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113121113122122311311222113111231133221121113122113121113221112131221123113111231121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321132132211322132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322123211211131211121332211231131122211311122122111312211213211312111322211231131122211311123113322112111331121113112221121113122113111231133221121113122113121113222123211211131211121332211213211321322113311213212312311211131122211213211321223112111311222112111312211312113221133221121113122113221321132132211231131122211311123113111213211221121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};
return arr[n];
}
}
- 时间复杂度O(1)
- 空间复杂度O(1)
思路二:递归
- 上面这种方式肯定不是要考察的点只能算取巧
- 依据前一个状态找到新的状态
- 这种很容易想到递归方法解决
- 我们只需要通过递归不断通过前一状态求解即可
public String countAndSay(int n) {
StringBuilder res = new StringBuilder();
//递归
if (n == 1) {
return "1";
}
int pre = 0;
int cur = 1;
String str = countAndSay(n - 1);
for (cur = 1; cur < str.length(); cur++) {
if (str.charAt(pre) != str.charAt(cur)) {// 如果碰到当前字符与前面紧邻的字符不等则更新此次结果
int count = cur - pre;
res.append(count).append(str.charAt(pre));
pre = cur;
}
}
if (pre != cur) {
int count = cur - pre;
res.append(count).append(str.charAt(pre));
}
return res.toString();
}
- 时间复杂度O(n)
- 空间复杂度O(n)