[Swift]LeetCode1092. 最短公共超序列 | Shortest Common Supersequence

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Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences.  If multiple answers exist, you may return any of them.

(A string S is a subsequence of string T if deleting some number of characters from T (possibly 0, and the characters are chosen anywhere from T) results in the string S.) 

Example 1:

Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation: 
str1 = "abac" is a substring of "cabac" because we can delete the first "c".
str2 = "cab" is a substring of "cabac" because we can delete the last "ac".
The answer provided is the shortest such string that satisfies these properties. 

Note:

1. 1 <= str1.length, str2.length <= 1000
2. str1 and str2 consist of lowercase English letters.

---

给出两个字符串 str1 和 str2，返回同时以 str1 和 str2 作为子序列的最短字符串。如果答案不止一个，则可以返回满足条件的任意一个答案。

（如果从字符串 T 中删除一些字符（也可能不删除，并且选出的这些字符可以位于 T 中的 任意位置），可以得到字符串 S，那么 S 就是 T 的子序列） 

示例：

输入：str1 = "abac", str2 = "cab"
输出："cabac"
解释：
str1 = "abac" 是 "cabac" 的一个子串，因为我们可以删去 "cabac" 的第一个 "c"得到 "abac"。 
str2 = "cab" 是 "cabac" 的一个子串，因为我们可以删去 "cabac" 末尾的 "ac" 得到 "cab"。
最终我们给出的答案是满足上述属性的最短字符串。 

提示：

1. 1 <= str1.length, str2.length <= 1000
2. str1 和 str2 都由小写英文字母组成。

---

212ms

class Solution {
    func shortestCommonSupersequence(_ str1: String, _ str2: String) -> String {
        let matrix = lcsLength(str1, str2)
        let comSubseq = backtrack(matrix, str1, str2)
        var chars1 = Array(str1)
        var chars2 = Array(str2)
        var i = 0 
        var j = 0
        var result = ""
        for c in comSubseq {
            while chars1[i] != c {
                result += String(chars1[i])
                i += 1
            }
            i += 1
            while chars2[j] != c {
                result += String(chars2[j])
                j += 1
            }
            j += 1
            result += String(c)
        }
        while i < chars1.count {
            result += String(chars1[i])
            i += 1
        }
        while j < chars2.count {
            result += String(chars2[j])
            j += 1
        }
        return result
    }
    
    
    fileprivate func lcsLength(_ str1: String, _ str2: String) -> [[Int]] {

        var matrix = [[Int]](repeating: [Int](repeating: 0, count: str2.count+1), count: str1.count+1)
        for (i, str1Char) in str1.enumerated() {
            for (j, str2Char) in str2.enumerated() {
                if str2Char == str1Char {
                    matrix[i+1][j+1] = matrix[i][j] + 1
                } else {
                    matrix[i+1][j+1] = max(matrix[i][j+1], matrix[i+1][j])
                }
            }
        }
        return matrix
    }
        
        
    fileprivate func backtrack(_ matrix: [[Int]], _ str1:String, _ str2: String) -> String {
        var i = str1.count
        var j = str2.count
        var charInSequence = str1.endIndex
        var lcs = String()
        while i >= 1 && j >= 1 {
            if matrix[i][j] == matrix[i][j - 1] {
                j -= 1
            } else if matrix[i][j] == matrix[i - 1][j] {
                i -= 1
                charInSequence = str1.index(before: charInSequence)
            } else {
                i -= 1
                j -= 1
                charInSequence = str1.index(before: charInSequence)
                lcs.append(str1[charInSequence])
            }
        }
        return String(lcs.reversed())
    }
}

---

Runtime: 528 ms

Memory Usage: 28.8 MB

class Solution {
    var dp:[[Int]] = [[Int]](repeating:[Int](repeating:-1,count:1010),count:1010)
    var s1:[Character] = [Character]()
    var s2:[Character] = [Character]()
    
    func shortestCommonSupersequence(_ str1: String, _ str2: String) -> String {
        self.s1 = Array(str1)
        self.s2 = Array(str2)
        var r:String = String()
        rec(0, 0, &r)
        return r
    }
    
    func dfs(_ x:Int,_ y:Int) -> Int
    {
        if x == s1.count && y == s2.count
        {
            return 0
        }
        if dp[x][y] != -1 {return dp[x][y]}
        var ans:Int = Int(1e9)
        if x < s1.count
        {
            ans = min(ans, dfs(x+1, y) + 1)
        }
        if y < s2.count
        {
            ans = min(ans, dfs(x, y+1) + 1)
        }
        if x < s1.count && y < s2.count && s1[x] == s2[y]
        {
            ans = min(ans, dfs(x+1, y+1) + 1)
        }
        dp[x][y] = ans
        return ans
    }
    
    func rec(_ x:Int,_ y:Int,_ r:inout String)
    {
        if x == s1.count && y == s2.count
        {
            return
        }
        if x < s1.count
        {
            if dfs(x+1, y) + 1 == dfs(x,y)
            {
                r.append(s1[x])
                rec(x+1, y, &r)
                return
            }
        }
        if y < s2.count
        {
            if dfs(x, y+1) + 1 == dfs(x,y)
            {
                r.append(s2[y])
                rec(x, y+1, &r)
                return
            }
        }
        if x < s1.count && y < s2.count && s1[x] == s2[y]
        {
            if dfs(x+1, y+1) + 1 == dfs(x,y)
            {
                r.append(s2[y])
                rec(x+1, y+1, &r)
                return
            }
        }
        return
    }
}