[Swift]LeetCode903. DI 序列的有效排列 | Valid Permutations for DI Sequence

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We are given S, a length n string of characters from the set {'D', 'I'}. (These letters stand for "decreasing" and "increasing".)

A valid permutation is a permutation P[0], P[1], ..., P[n] of integers {0, 1, ..., n}, such that for all i:

• If S[i] == 'D', then P[i] > P[i+1], and;
• If S[i] == 'I', then P[i] < P[i+1].

How many valid permutations are there?  Since the answer may be large, return your answer modulo 10^9 + 7. 

Example 1:

Input: "DID"
Output: 5
Explanation: 
The 5 valid permutations of (0, 1, 2, 3) are:
(1, 0, 3, 2)
(2, 0, 3, 1)
(2, 1, 3, 0)
(3, 0, 2, 1)
(3, 1, 2, 0) 

Note:

1. 1 <= S.length <= 200
2. S consists only of characters from the set {'D', 'I'}.

---

我们给出 S，一个源于 {'D', 'I'} 的长度为 n 的字符串 。（这些字母代表 “减少” 和 “增加”。）
有效排列 是对整数 {0, 1, ..., n} 的一个排列 P[0], P[1], ..., P[n]，使得对所有的 i：

• 如果 S[i] == 'D'，那么 P[i] > P[i+1]，以及；
• 如果 S[i] == 'I'，那么 P[i] < P[i+1]。

有多少个有效排列？因为答案可能很大，所以请返回你的答案模 10^9 + 7. 

示例：

输入："DID"
输出：5
解释：
(0, 1, 2, 3) 的五个有效排列是：
(1, 0, 3, 2)
(2, 0, 3, 1)
(2, 1, 3, 0)
(3, 0, 2, 1)
(3, 1, 2, 0) 

提示：

1. 1 <= S.length <= 200
2. S 仅由集合 {'D', 'I'} 中的字符组成。

---

Runtime: 16 ms

Memory Usage: 19.8 MB

class Solution {
    func numPermsDISequence(_ S: String) -> Int {
        var n:Int = S.count
        var mod:Int = Int(1e9 + 7)
        var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n + 1),count:n + 1)
        for j in 0...n
        {
            dp[0][j] = 1
        }
        let arrS:[Character] = Array(S)
        for i in 0..<n
        {
            if arrS[i] == "I"
            {
                var j:Int = 0
                var cur:Int = 0
                while(j < n - i)
                {                     
                    cur = (cur + dp[i][j]) % mod
                    dp[i + 1][j] = cur
                    j += 1
                }
            }
            else
            {
                var j:Int = n - i - 1
                var cur:Int = 0
                while(j >= 0)
                {
                    cur = (cur + dp[i][j + 1]) % mod
                    dp[i + 1][j] = cur
                    j -= 1
                }
            }
        }
        return dp[n][0]
    }
}