[Swift]LeetCode730. 统计不同回文子字符串 | Count Different Palindromic Subsequences

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Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7.

A subsequence of a string S is obtained by deleting 0 or more characters from S.

A sequence is palindromic if it is equal to the sequence reversed.

Two sequences A_1, A_2, ... and B_1, B_2, ... are different if there is some i for which A_i != B_i.

Example 1:

Input: 
S = 'bccb'
Output: 6
Explanation: 
The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'.
Note that 'bcb' is counted only once, even though it occurs twice. 

Example 2:

Input: 
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
Output: 104860361
Explanation: 
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7. 

Note:

• The length of S will be in the range [1, 1000].
• Each character S[i] will be in the set {'a', 'b', 'c', 'd'}.

---

给定一个字符串 S，找出 S 中不同的非空回文子序列个数，并返回该数字与 10^9 + 7 的模。

通过从 S 中删除 0 个或多个字符来获得子字符序列。

如果一个字符序列与它反转后的字符序列一致，那么它是回文字符序列。

如果对于某个  i，A_i != B_i，那么 A_1, A_2, ... 和 B_1, B_2, ... 这两个字符序列是不同的。 

示例 1：

输入：
S = 'bccb'
输出：6
解释：
6 个不同的非空回文子字符序列分别为：'b', 'c', 'bb', 'cc', 'bcb', 'bccb'。
注意：'bcb' 虽然出现两次但仅计数一次。

示例 2：

输入：
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
输出：104860361
解释：
共有 3104860382 个不同的非空回文子字符序列，对 10^9 + 7 取模为 104860361。 

提示：

• 字符串 S 的长度将在[1, 1000]范围内。
• 每个字符 S[i] 将会是集合 {'a', 'b', 'c', 'd'} 中的某一个。

---

Runtime: 760 ms

Memory Usage: 25.9 MB

class Solution {
    func countPalindromicSubsequences(_ S: String) -> Int {
        var arr:[Character] = Array(S)
        var n:Int = S.count
        var M:Int = Int(1e9 + 7)
        var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n),count:n)
        for i in 0..<n
        {
            dp[i][i] = 1
        }
        for len in 1..<n
        {
            for i in 0..<(n - len)
            {
                var j:Int = i + len
                if arr[i] == arr[j]
                {
                    var left:Int = i + 1
                    var right:Int = j - 1
                    while (left <= right && arr[left] != arr[i])
                    {
                        left += 1
                    }
                    while (left <= right && arr[right] != arr[i])
                    {
                        right -= 1
                    }
                    if left > right
                    {
                        dp[i][j] = dp[i + 1][j - 1] * 2 + 2
                    }
                    else if left == right
                    {
                        dp[i][j] = dp[i + 1][j - 1] * 2 + 1
                    }
                    else
                    {
                        dp[i][j] = dp[i + 1][j - 1] * 2 - dp[left + 1][right - 1]
                    }
                }
                else
                {
                    dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1]
                }
                dp[i][j] = (dp[i][j] < 0) ? dp[i][j] + M : dp[i][j] % M
            }
        }
        return dp[0][n - 1]
    }
}