[Swift]LeetCode132. 分割回文串 II | Palindrome Partitioning II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/9963223.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

Example:

Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

---

给定一个字符串 s，将 s 分割成一些子串，使每个子串都是回文串。

返回符合要求的最少分割次数。

示例:

输入: "aab"
输出: 1
解释: 进行一次分割就可将 s 分割成 ["aa","b"] 这样两个回文子串。

---

148ms

class Solution {
    func minCut(_ s: String) -> Int {
        var isPalindrome = Array(repeating: Array(repeating: false, count: s.count), count: s.count)
        var res = Array(repeating: Int.max, count: s.count)
        let sArray = Array(s)
        
        for i in 0..<s.count {
            for j in 0..<i + 1 {
                if sArray[j] == sArray[i] && (i - j < 2 || isPalindrome[j + 1][i - 1]) {
                    res[i] = j == 0 ? 0 : min(res[i], res[j - 1] + 1)
                    isPalindrome[j][i] = true
                }
            }
        }
        
        return res[res.count - 1]
    }
}

---

148ms

class Solution {
    func minCut(_ s: String) -> Int {
        if s.isEmpty {
            return 0
        }
        
        let count = s.count
        var nums = Array(repeating: 0, count: count+1)
        for i in 0...count {
            nums[i] = count - i - 1
        }
        var p = Array(repeating: Array(repeating: false, count: count), count: count)
        let sArr = Array(s)
        
        for i in stride(from: count-1, to: -1, by: -1) {
            for j in i..<count {
                if sArr[i] == sArr[j] && (j-i<=1 || p[i+1][j-1]) {
                    p[i][j] = true
                    nums[i] = min(nums[i], nums[j+1] + 1)
                }
            }
        }
        
        
        return nums[0]
    }
}

---

156ms

class Solution {
    func minCut(_ s: String) -> Int {

        var n = s.count
        guard n > 1 else { return 0 }
        let s = Array(s)
        var p = Array(repeating: Array(repeating: false, count: n), count: n )
        var dp = [Int]()
        for i in 0...n { dp.append(n-i-1) }
        for i in (0...n-1).reversed() {
            for j in i...n-1 {
                if s[i] == s[j], (j-i <= 1 || p[i+1][j-1]) {
                    p[i][j] = true
                    dp[i] = min(dp[i], dp[j+1]+1)
                }
            }
        }
        return dp[0]
    }
}

---

172ms

class Solution {
    func minCut(_ s: String) -> Int {
        var res = [Int](repeating: Int.max, count: s.count)
        var isPalindrome = Array(repeating: Array(repeating: false, count: s.count), count: s.count)
        let sArray = Array(s)
        
        for i in 0..<s.count {
            for j in 0..<i + 1 {
                if sArray[i] == sArray[j] && (i - j < 2 || isPalindrome[j + 1][i - 1]) {
                    res[i] = j == 0 ? 0 : min(res[i], res[j - 1] + 1)
                    isPalindrome[j][i] = true
                }
            }
        }
        
        return res[res.count - 1]
    }
}