[Swift]LeetCode524. 通过删除字母匹配到字典里最长单词 | Longest Word in Dictionary through Deleting

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Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.

Example 1:

Input:
s = "abpcplea", d = ["ale","apple","monkey","plea"]

Output: 
"apple"

Example 2:

Input:
s = "abpcplea", d = ["a","b","c"]

Output: 
"a"

Note:

1. All the strings in the input will only contain lower-case letters.
2. The size of the dictionary won't exceed 1,000.
3. The length of all the strings in the input won't exceed 1,000.

---

给定一个字符串和一个字符串字典，找到字典里面最长的字符串，该字符串可以通过删除给定字符串的某些字符来得到。如果答案不止一个，返回长度最长且字典顺序最小的字符串。如果答案不存在，则返回空字符串。

示例 1:

输入:
s = "abpcplea", d = ["ale","apple","monkey","plea"]

输出: 
"apple"

示例 2:

输入:
s = "abpcplea", d = ["a","b","c"]

输出: 
"a"

说明:

1. 所有输入的字符串只包含小写字母。
2. 字典的大小不会超过 1000。
3. 所有输入的字符串长度不会超过 1000。

---

340ms

class Solution {
    func findLongestWord(_ s: String, _ d: [String]) -> String {
    var result : [Character] = []
    let s = Array(s)
    for str in d {
        var str = Array(str)
        if str.count < result.count{
            continue
        }
        
        if canBeFoundFrom(s: s, with: Array(str)){
            if str.count > result.count{
                result = str
            }else if str.count == result.count{
                result = lexicoOrder(result, str)
            }
            
        }
    }
    
    return String(result)
}

func lexicoOrder(_ first : [Character], _ second : [Character])->[Character]{
    
    var firstIndex : Int = 0
    var secondIndex : Int = 0
    
    while firstIndex < first.count{
        if first[firstIndex] == second[secondIndex]{
            firstIndex += 1
            secondIndex += 1
        }else if first[firstIndex] < second[secondIndex]{
            return first
        }else{
            return second
        }
    }
    return first
}

func canBeFoundFrom(s : [Character], with d : [Character])->Bool{
    
    
    var sIndex : Int = 0
    var dIndex : Int = 0
    while sIndex < s.count{
        if d[dIndex] == s[sIndex]{
            dIndex += 1
        }
        sIndex += 1
        if dIndex == d.count{
            return true
        }
    }
    
    return dIndex == d.count
  }
}

---

532ms

class Solution {
     func findLongestWord(_ s: String, _ d: [String]) -> String {
        if d.count == 0 {
            return ""
        }

        var long = ""
        for items in d {
            let i = long.count
            let j = items.count
            if i > j || (i == j && long < items) {
                continue
            }
            if isValid(s: s, d: items) {
                long = items
            }
        }
        
        return long
    }
    
    fileprivate func isValid(s: String, d: String) -> Bool {
        var i = 0
        var j = 0
        var source = Array(s)
        var target = Array(d)
        var result = ""
        while i < source.count && j < target.count {
            if source[i] == target[j] {
                j += 1
                result.append(source[i])
            }
            i += 1
        }
        return j == target.count
    }
}

---

840ms

class Solution {
    func findLongestWord(_ s: String, _ d: [String]) -> String {
    
    let sortedD = d.sorted { (first, second) -> Bool in
        if first.count > second.count{
            return true
        }else if first.count == second.count{
            return first < second
        }else{
            return false
        }
    }
    
    for str in sortedD{
        if possibleToForm(s, str: str){
            return str
        }
    }
    return ""
}

func possibleToForm(_ base : String, str : String)->Bool{
    guard  base.count >= str.count else {
        return false
    }
    
    var baseIndex : String.Index = base.startIndex
    var strIndex : String.Index = str.startIndex
    
    while baseIndex != base.endIndex && strIndex != str.endIndex{
        if str[strIndex] == base[baseIndex]{
            strIndex = str.index(after: strIndex)
        }
         baseIndex = base.index(after: baseIndex)
    }
    
    return strIndex == str.endIndex
  }
}

---

Runtime: 7968 ms

Memory Usage: 20.7 MB

class Solution {
    func findLongestWord(_ s: String, _ d: [String]) -> String {
        var res:String = String()        
        for str in d
        {
            var arr:[Character] = Array(str)
            var i:Int = 0
            for c in s.characters
            {
                if i < str.count && c == arr[i]
                {
                    i += 1
                }
            }
            if i == str.count && str.count >= res.count
            {
                if str.count > res.count || str < res
                {
                    res = str
                }
            }   
        }
        return res
    }
}