• [Swift]LeetCode791. 自定义字符串排序 | Custom Sort String


    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤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/10545744.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!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    S and T are strings composed of lowercase letters. In S, no letter occurs more than once.

    S was sorted in some custom order previously. We want to permute the characters of T so that they match the order that S was sorted. More specifically, if x occurs before y in S, then x should occur before y in the returned string.

    Return any permutation of T (as a string) that satisfies this property.

    Example :
    Input: 
    S = "cba"
    T = "abcd"
    Output: "cbad"
    Explanation: 
    "a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a". 
    Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.

    Note:

    • S has length at most 26, and no character is repeated in S.
    • T has length at most 200.
    • S and T consist of lowercase letters only.

    字符串S和 T 只包含小写字符。在S中,所有字符只会出现一次。

    S 已经根据某种规则进行了排序。我们要根据S中的字符顺序对T进行排序。更具体地说,如果Sxy之前出现,那么返回的字符串中x也应出现在y之前。

    返回任意一种符合条件的字符串T

    示例:
    输入:
    S = "cba"
    T = "abcd"
    输出: "cbad"
    解释: 
    S中出现了字符 "a", "b", "c", 所以 "a", "b", "c" 的顺序应该是 "c", "b", "a". 
    由于 "d" 没有在S中出现, 它可以放在T的任意位置. "dcba", "cdba", "cbda" 都是合法的输出。
    

    注意:

    • S的最大长度为26,其中没有重复的字符。
    • T的最大长度为200
    • ST只包含小写字符。

    Runtime: 8 ms
    Memory Usage: 19.8 MB
     1 class Solution {
     2     func customSortString(_ S: String, _ T: String) -> String {
     3         var res:String = String()
     4         var m:[Character:Int] = [Character:Int]()
     5         for c in T
     6         {
     7             m[c,default:0] += 1
     8         }
     9         for c in S
    10         {
    11             let num:Int = m[c,default:0]
    12             for _ in 0..<num
    13             {
    14                 res.append(c)
    15             }
    16             m[c,default:0] = 0
    17         }
    18         for (key,val) in m
    19         {
    20             for _ in 0..<val
    21             {
    22                 res.append(key)
    23             }
    24         }
    25         return res
    26     }
    27 }

    8ms

     1 class Solution {
     2     func customSortString(_ S: String, _ T: String) -> String {
     3         var tMap: Dictionary<Character, Int> = [:]
     4         for t in T {
     5             if let v = tMap[t] {
     6                 tMap[t] = v + 1
     7             } else {
     8                 tMap[t] = 1
     9             }
    10         }
    11         
    12         var result = ""
    13         for s in S {
    14             if let v = tMap[s] {
    15                 result.append(String(repeating: s, count: v))
    16                 tMap.removeValue(forKey: s)
    17             }
    18         }
    19         
    20         for k in tMap.keys {
    21             result.append(String(repeating: k, count: tMap[k]!))
    22         }
    23         
    24         return result
    25     }
    26 }

    24ms

    1 class Solution {
    2     func customSortString(_ S: String, _ T: String) -> String {
    3         let map = S.enumerated().reduce(into: [:]) { $0[$1.1, default: 0] = $1.0 }
    4         let character = T.sorted(by: { map[$0, default: Int.max] < map[$1, default: Int.max] })
    5         return String(character)
    6     }
    7 }

    32ms

     1 class Solution {
     2     static func relativeCode(_ c: Character) -> Int {
     3         return Int(c.unicodeScalars.first!.value - Character("a").unicodeScalars.first!.value)
     4     }
     5     func customSortString(_ S: String, _ T: String) -> String {
     6         var order = [Int](repeating: -1, count: 26)
     7         for (i, s) in S.enumerated() {
     8             order[Solution.relativeCode(s)] = i
     9         }
    10         
    11         return String(T.sorted {
    12             return order[Solution.relativeCode($0)] < order[Solution.relativeCode($1)]
    13         })
    14     }
    15 }
  • 相关阅读:
    HDU2363 最短路+贪心
    stl-----map去重,排序,计数
    STL------sort三种比较算子定义
    栈------表达式求值
    踩水坑系列一
    第一周 动态规划Dynamic Programming(一)
    模拟递归回溯贪心专题入门
    HDU1013,1163 ,2035九余数定理 快速幂取模
    HDU1005 找规律 or 循环点 or 矩阵快速幂
    入门基础常识
  • 原文地址:https://www.cnblogs.com/strengthen/p/10545744.html
Copyright © 2020-2023  润新知