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

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.

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。

 1 class Solution {
 2     func minDistance(_ word1: String, _ word2: String) -> Int {
 3         if word1.count < word2.count {
 4             return minDistance(word2, word1)
 5         }
 6         var cache = Array(repeating: Array(repeating: -1, count: word2.count), count: word1.count)
 7         return word1.count + word2.count - lcs(Array(word1), Array(word2), word1.count - 1, word2.count - 1, &cache)*2
 8     }
 9
10     func lcs(_ word1:[Character], _ word2:[Character], _ i:Int, _ j:Int, _ cache:inout [[Int]]) -> Int {
11         var maxLenght = 0
12         if i < 0 || j < 0 {
13             return maxLenght
14         }
15         if cache[i][j] != -1 {
16             return cache[i][j]
17         }
18         if word1[i] == word2[j] {
19             maxLenght = max(maxLenght, lcs(word1, word2, i - 1, j - 1, &cache) + 1)
20         } else {
21             maxLenght = max(maxLenght, lcs(word1, word2, i, j - 1, &cache), lcs(word1, word2, i - 1, j, &cache))
22         }
23         cache[i][j] = maxLenght
24         return maxLenght
25     }
26 }

 1 class Solution {
 2     func minDistance(_ word1: String, _ word2: String) -> Int {
 3         if word1.count == 0 {
 4             return word2.count
 5         }
 6         if word2.count == 0 {
 7             return word1.count
 8         }
 9         let len = longestCommonSequence(Array(word1), Array(word2))
10         return word1.count + word2.count - 2 * len
11     }
12
13     func longestCommonSequence(_ string1: [Character], _ string2: [Character]) -> Int {
14         let m = string1.count
15         let n = string2.count
16         var dp = [[Int]]()
17         for _ in 0 ..< m + 1 {
18             let array = [Int].init(repeating: 0, count: n + 1)
19             dp.append(array)
20         }
21
22         for i in 0 ..< m {
23             for j in 0 ..< n {
24                 if string1[i] == string2[j] {
25                     dp[i+1][j+1] = dp[i][j] + 1
26                 } else {
27                     dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
28                 }
29             }
30         }
31         return dp[m][n]
32     }
33 }

 1 class Solution {
 2     func minDistance(_ word1: String, _ word2: String) -> Int {
 3     var word1 = Array(word1)
 4     var word2 = Array(word2)
 5     var dp = [[Int]](repeating: [Int](repeating: 0, count: word2.count + 1), count: word1.count + 1)
 6
 7     for row in 0..<dp.count{
 8         dp[row][0] = row
 9     }
10
11     for column in 0..<dp[0].count{
12         dp[0][column] = column
13     }
14
15     for row in 1..<dp.count{
16         for column in 1..<dp[row].count{
17             if word1[row-1] == word2[column-1]{
18                 dp[row][column] = dp[row - 1][column - 1]
19             }else{
20                 dp[row][column] = min(dp[row - 1][column], dp[row][column - 1]) + 1
21             }
22         }
23     }
24
25     return dp.last!.last!
26   }
27 }

 1 class Solution {
 2     func minDistance(_ word1: String, _ word2: String) -> Int {
 3         let word1 = Array(word1)
 4         let word2 = Array(word2)
 5         let len1 = word1.count
 6         let len2 = word2.count
 7         guard len1 != 0 else { return len2 }
 8         guard len2 != 0 else { return len1 }
 9
10         var dp: [[Int]] = Array(repeating: Array(repeating: 0, count: len2 + 1), count: len1 + 1)
11         for i in 0 ... len1 {
12             dp[i][0] = i
13         }
14
15         for i in 0 ... len2 {
16             dp[0][i] = i
17         }
18
19         for i in 1 ... len1 {
20             for j in 1 ... len2 {
21                 if word1[i - 1] == word2[j - 1] {
22                     dp[i][j] = dp[i - 1][j - 1]
23                 } else {
24                     dp[i][j] = min(dp[i - 1][j - 1] + 2, min(dp[i][j - 1] + 1, dp[i - 1][j] + 1))
25                 }
26             }
27         }
28
29         return dp[len1][len2]
30     }
31 }

Memory Usage: 20.4 MB

 1 class Solution {
 2     func minDistance(_ word1: String, _ word2: String) -> Int {
 3         var n1:Int = word1.count
 4         var n2:Int = word2.count
 5         var memo:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n2 + 1),count:n1 + 1)
 6         return helper(word1, word2, 0, 0, &memo)
 7     }
 8
 9     func helper(_ word1: String, _ word2: String,_ p1:Int,_ p2:Int,_ memo:inout [[Int]]) -> Int
10     {
11         if memo[p1][p2] != 0
12         {
13             return memo[p1][p2]
14         }
15         var n1:Int = word1.count
16         var n2:Int = word2.count
17         if p1 == n1 || p2 == n2
18         {
19             return n1 - p1 + n2 - p2
20         }
21         if word1[p1] == word2[p2]
22         {
23             memo[p1][p2] = helper(word1, word2, p1 + 1, p2 + 1, &memo)
24         }
25         else
26         {
27              memo[p1][p2] = 1 + min(helper(word1, word2, p1 + 1, p2, &memo), helper(word1, word2, p1, p2 + 1, &memo))
28         }
29         return memo[p1][p2]
30     }
31 }
32
33 extension String {
34     //subscript函数可以检索数组中的值
35     //直接按照索引方式截取指定索引的字符
36     subscript (_ i: Int) -> Character {
37         //读取字符
38         get {return self[index(startIndex, offsetBy: i)]}
39     }
40 }

原文地址：https://www.cnblogs.com/strengthen/p/10420808.html