zoukankan      html  css  js  c++  java
  • [Swift]LeetCode1087. 字母切换 | Permutation of Letters

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤微信公众号:山青咏芝(shanqingyongzhi)
    ➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
    ➤GitHub地址:https://github.com/strengthen/LeetCode
    ➤原文地址:https://www.cnblogs.com/strengthen/p/11031602.html 
    ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
    ➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    A string S represents a list of words.

    Each letter in the word has 1 or more options.  If there is one option, the letter is represented as is.  If there is more than one option, then curly braces delimit the options.  For example, "{a,b,c}" represents options ["a", "b", "c"].

    For example, "{a,b,c}d{e,f}" represents the list ["ade", "adf", "bde", "bdf", "cde", "cdf"].

    Return all words that can be formed in this manner, in lexicographical order. 

    Example 1:

    Input: "{a,b}c{d,e}f"
    Output: ["acdf","acef","bcdf","bcef"]
    

    Example 2:

    Input: "abcd"
    Output: ["abcd"] 

    Note:

    1. 1 <= S.length <= 50
    2. There are no nested curly brackets.
    3. All characters inside a pair of consecutive opening and ending curly brackets are different.

    我们用一个特殊的字符串 S 来表示一份单词列表,之所以能展开成为一个列表,是因为这个字符串 S 中存在一个叫做「选项」的概念:

    单词中的每个字母可能只有一个选项或存在多个备选项。如果只有一个选项,那么该字母按原样表示。

    如果存在多个选项,就会以花括号包裹来表示这些选项(使它们与其他字母分隔开),例如 "{a,b,c}" 表示 ["a", "b", "c"]

    例子:"{a,b,c}d{e,f}" 可以表示单词列表 ["ade", "adf", "bde", "bdf", "cde", "cdf"]

    请你按字典顺序,返回所有以这种方式形成的单词。 

    示例 1:

    输入:"{a,b}c{d,e}f"
    输出:["acdf","acef","bcdf","bcef"]
    

    示例 2:

    输入:"abcd"
    输出:["abcd"] 

    提示:

    1. 1 <= S.length <= 50
    2. 你可以假设题目中不存在嵌套的花括号
    3. 在一对连续的花括号(开花括号与闭花括号)之间的所有字母都不会相同

    76 ms

     1 class Solution {
     2     func permute(_ S: String) -> [String] {
     3         var res:[String] = [String]()
     4         res.append(String())
     5         var cs:[Character] = Array(S)
     6         let l:Int = cs.count
     7         var i:Int = 0
     8         while(i < l)
     9         {
    10             if cs[i] == "{"
    11             {
    12                 var t:String = String()
    13                 i += 1
    14                 while(cs[i] != "}")
    15                 {
    16                     t.append(cs[i])
    17                     i += 1
    18                 }
    19                 let x:[String] = t.split{$0 == ","}.map(String.init)
    20                 var res1:[String] = [String]()
    21                 for g in x
    22                 {
    23                     for temp in res
    24                     {
    25                         res1.append(temp + g)
    26                     }
    27                 }
    28                 res = res1
    29                 
    30             }
    31             else
    32             {
    33                 var res1:[String] = [String]()
    34                 for temp in res
    35                 {
    36                     var str = temp
    37                     str.append(cs[i])
    38                     res1.append(str)
    39                 }
    40                 res = res1
    41             }
    42             i += 1
    43         }
    44         res.sort()
    45         return res
    46     }
    47 }
  • 相关阅读:
    UPC 5130 Concerts
    poj 1079 Calendar Game
    2018 ACM-ICPC 中国大学生程序设计竞赛线上赛
    CF932E
    浅谈Tarjan算法
    拉格朗日差值
    扩展欧几里得算法(exgcd)
    欧拉定理
    莫比乌斯反演
    除法分块
  • 原文地址:https://www.cnblogs.com/strengthen/p/11031602.html
Copyright © 2011-2022 走看看