[Swift]LeetCode583. 两个字符串的删除操作 | Delete Operation for Two Strings

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Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea". 

Note:

1. The length of given words won't exceed 500.
2. Characters in given words can only be lower-case letters.

---

给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数，每步可以删除任意一个字符串中的一个字符。

示例 1:

输入: "sea", "eat"
输出: 2
解释: 第一步将"sea"变为"ea"，第二步将"eat"变为"ea"

说明:

1. 给定单词的长度不超过500。
2. 给定单词中的字符只含有小写字母。

---

124ms

class Solution {
    func minDistance(_ word1: String, _ word2: String) -> Int {
        if word1.count < word2.count {
            return minDistance(word2, word1)
        }
        var cache = Array(repeating: Array(repeating: -1, count: word2.count), count: word1.count)
        return word1.count + word2.count - lcs(Array(word1), Array(word2), word1.count - 1, word2.count - 1, &cache)*2
    }
    
    func lcs(_ word1:[Character], _ word2:[Character], _ i:Int, _ j:Int, _ cache:inout [[Int]]) -> Int {
        var maxLenght = 0
        if i < 0 || j < 0 {
            return maxLenght
        }
        if cache[i][j] != -1 {
            return cache[i][j]
        }
        if word1[i] == word2[j] {
            maxLenght = max(maxLenght, lcs(word1, word2, i - 1, j - 1, &cache) + 1)
        } else {
            maxLenght = max(maxLenght, lcs(word1, word2, i, j - 1, &cache), lcs(word1, word2, i - 1, j, &cache))
        }
        cache[i][j] = maxLenght
        return maxLenght
    }
}

---

132ms

class Solution {
    func minDistance(_ word1: String, _ word2: String) -> Int {
        if word1.count == 0 {
            return word2.count
        }
        if word2.count == 0 {
            return word1.count
        }
        let len = longestCommonSequence(Array(word1), Array(word2))
        return word1.count + word2.count - 2 * len
    }
    
    func longestCommonSequence(_ string1: [Character], _ string2: [Character]) -> Int {
        let m = string1.count
        let n = string2.count
        var dp = [[Int]]()
        for _ in 0 ..< m + 1 {
            let array = [Int].init(repeating: 0, count: n + 1)
            dp.append(array)
        }
        
        for i in 0 ..< m {
            for j in 0 ..< n {
                if string1[i] == string2[j] {
                    dp[i+1][j+1] = dp[i][j] + 1
                } else {
                    dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
                }
            }
        }
        return dp[m][n]
    }
}

---

136ms

class Solution {
    func minDistance(_ word1: String, _ word2: String) -> Int {
    var word1 = Array(word1)
    var word2 = Array(word2)
    var dp = [[Int]](repeating: [Int](repeating: 0, count: word2.count + 1), count: word1.count + 1)
    
    for row in 0..<dp.count{
        dp[row][0] = row
    }
    
    for column in 0..<dp[0].count{
        dp[0][column] = column
    }
    
    for row in 1..<dp.count{
        for column in 1..<dp[row].count{
            if word1[row-1] == word2[column-1]{
                dp[row][column] = dp[row - 1][column - 1]
            }else{
                dp[row][column] = min(dp[row - 1][column], dp[row][column - 1]) + 1
            }
        }
    }
    
    return dp.last!.last!
  }
}

---

156ms

class Solution {
    func minDistance(_ word1: String, _ word2: String) -> Int {
        let word1 = Array(word1)
        let word2 = Array(word2)
        let len1 = word1.count
        let len2 = word2.count
        guard len1 != 0 else { return len2 }
        guard len2 != 0 else { return len1 }
        
        var dp: [[Int]] = Array(repeating: Array(repeating: 0, count: len2 + 1), count: len1 + 1)
        for i in 0 ... len1 {
            dp[i][0] = i
        }
        
        for i in 0 ... len2 {
            dp[0][i] = i
        }
        
        for i in 1 ... len1 {
            for j in 1 ... len2 {
                if word1[i - 1] == word2[j - 1] {
                    dp[i][j] = dp[i - 1][j - 1]
                } else {
                    dp[i][j] = min(dp[i - 1][j - 1] + 2, min(dp[i][j - 1] + 1, dp[i - 1][j] + 1))
                }
            }
        }
        
        return dp[len1][len2]
    }
}

---

Runtime: 732 ms

Memory Usage: 20.4 MB

class Solution {
    func minDistance(_ word1: String, _ word2: String) -> Int {
        var n1:Int = word1.count
        var n2:Int = word2.count
        var memo:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n2 + 1),count:n1 + 1)
        return helper(word1, word2, 0, 0, &memo)
    }
    
    func helper(_ word1: String, _ word2: String,_ p1:Int,_ p2:Int,_ memo:inout [[Int]]) -> Int
    {
        if memo[p1][p2] != 0
        {
            return memo[p1][p2]
        }
        var n1:Int = word1.count
        var n2:Int = word2.count
        if p1 == n1 || p2 == n2
        {
            return n1 - p1 + n2 - p2            
        }
        if word1[p1] == word2[p2]
        {
            memo[p1][p2] = helper(word1, word2, p1 + 1, p2 + 1, &memo)
        }
        else
        {
             memo[p1][p2] = 1 + min(helper(word1, word2, p1 + 1, p2, &memo), helper(word1, word2, p1, p2 + 1, &memo))
        }
        return memo[p1][p2]
    }
}

extension String {        
    //subscript函数可以检索数组中的值
    //直接按照索引方式截取指定索引的字符
    subscript (_ i: Int) -> Character {
        //读取字符
        get {return self[index(startIndex, offsetBy: i)]}
    }
}