[Swift]LeetCode97. 交错字符串 | Interleaving 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/9937036.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

---

给定三个字符串 s1, s2, s3, 验证 s3 是否是由 s1 和 s2 交错组成的。

示例 1:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出: true

示例 2:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出: false

---

12ms

class Solution {
    func isInterleave(_ s1: String, _ s2: String, _ s3: String) -> Bool {
        var cache = [[Int]](repeating: [Int](repeating: -1, count: s2.count), count: s1.count)
        return is_Interleave(s1, 0, s2, 0, s3, 0, &cache)
    }

    fileprivate func is_Interleave(_ s1: String, _ i: Int, _ s2: String, _ j: Int, _ s3: String, _ k: Int, _ cache: inout [[Int]]) -> Bool {
        if i == s1.count {
            return s2[j..<s2.count] == s3[k..<s3.count]
        }
        if j == s2.count {
            return s1[i..<s1.count] == s3[k..<s3.count]
        }
        if cache[i][j] >= 0 {
            return cache[i][j] == 1 ? true : false
        }

        var ans = false

        if (s3[k] == s1[i] && is_Interleave(s1, i + 1, s2, j, s3, k + 1, &cache)) ||
            (s3[k] == s2[j] && is_Interleave(s1, i, s2, j + 1, s3, k + 1, &cache)) {
            ans = true
        }
        cache[i][j] = ans ? 1 : 0

        return ans
    }
}

extension String {
    subscript (i: Int) -> Character {
        return self[index(startIndex, offsetBy: i)]
    }
    
    subscript(_ range: CountableRange<Int>) -> String {
        let idx1 = index(startIndex, offsetBy: max(0, range.lowerBound))
        let idx2 = index(startIndex, offsetBy: min(self.count, range.upperBound))
        return String(self[idx1..<idx2])
    }
}

---

12ms

class Solution {
    func isInterleave(_ s1: String, _ s2: String, _ s3: String) -> Bool {
        if s1.count + s2.count != s3.count {
            return false
        }
        var dp: [[Bool]] = Array<Array<Bool>>(repeating: Array<Bool>(repeating: false, count: s2.count+1), count: s1.count+1)
        for i in 0 ..< s1.count + 1{
            for j in 0 ..< s2.count + 1{
                if i == 0 && j == 0 {
                    dp[0][0] = true
                } else if i == 0 {
                    dp[0][j] = dp[0][j-1] && (s2[j-1] == s3[j-1])
                } else if j == 0 {
                    dp[i][0] = dp[i-1][0] && (s1[i-1] == s3[i-1])
                } else {
                    dp[i][j] = (dp[i][j-1] && (s2[j-1] == s3[i+j-1])) || (dp[i-1][j] && (s1[i-1] == s3[i+j-1]))
                }
            }
        }
        return dp[s1.count][s2.count]
    }
}

extension String {
    subscript(index: Int) -> Character {
        return self[self.index(self.startIndex, offsetBy: index)]
    }
}