639 wrong answer

Author: aQua

Algorithms/0639.decode-ways-ii/README.md

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+# [639. Decode Ways II](https://leetcode.com/problems/decode-ways-ii/)
+
+## 题目
+A message containing letters from `A-Z` is being encoded to numbers using the following mapping way:
+
+```
+'A' -> 1
+'B' -> 2
+...
+'Z' -> 26
+```
+
+Beyond that, now the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9.
+
+Given the encoded message containing digits and the character '*', return the total number of ways to decode it.
+
+Also, since the answer may be very large, you should return the output mod 109 + 7.
+
+```
+Example 1:
+Input: "*"
+Output: 9
+Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".
+
+Example 2:
+Input: "1*"
+Output: 9 + 9 = 18
+```
+
+Note:
+1. The length of the input string will fit in range [1, 105].
+1. The input string will only contain the character '*' and digits '0' - '9'.
+
+## 解题思路
+
+见程序注释

Algorithms/0639.decode-ways-ii/decode-ways-ii.go

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+package Problem0639
+
+func numDecodings(s string) int {
+	n := len(s)
+	if n == 0 {
+		return 0
+	}
+
+	dp := make([]int, n+1)
+	// dp[i] 表示 s[:i+1] 的组成方式数
+	dp[0], dp[1] = 1, one(s[0:1])
+
+	for i := 2; i <= n; i++ {
+		w1, w2 := one(s[i-1:i]), two(s[i-2:i])
+		dp[i] = dp[i-1]*w1 + dp[i-2]*w2
+		if dp[i] == 0 {
+			// 子字符串 s[:i+1] 的组成方式数为 0
+			// 则，s 的组成方式数肯定也为 0
+			// 这时可以提前结束
+			return 0
+		}
+	}
+
+	return dp[n]
+}
+
+// 检查 s 是否为合格的单字符
+// len(s) == 1
+// 合格，  返回 1
+// 不合格， 返回 0
+// 注意：
+//     题目保证了 s 只含有数字和'*'
+func one(s string) int {
+	switch s[0] {
+	case '*':
+		// '*' 可以表示 1~9
+		return 9
+	case '0':
+		return 0
+	default:
+		return 1
+	}
+}
+
+// 检查 s 是否为合格的双字符
+// len(s) == 2
+// 合格，  返回 1
+// 不合格， 返回 0
+// 注意：
+//     题目保证了 s 只含有数字和'*'
+func two(s string) int {
+	switch s[0] {
+	case '*':
+		if s[1] == '*' {
+			return 15
+		}
+		if '1' <= s[1] && s[1] <= '6' {
+			return 2
+		}
+		return 1
+	case '1':
+		if s[1] == '*' {
+			return 9
+		}
+		return 1
+	case '2':
+		if s[1] == '*' {
+			return 6
+		}
+		if s[1] == '7' || s[1] == '8' || s[1] == '9' {
+			return 0
+		}
+		return 1
+	default:
+		// '3' <= s[0] && s[0] <= '9'，或
+		// s[0] == '0'
+		return 0
+	}
+}