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:
- The length of the input string will fit in range [1, 105].
- The input string will only contain the character '*' and digits '0' - '9'.
91. Decode Ways 的拓展，这次字符串里面可能含有'*'， 它可以1~9中的任何一个，求总的解码方法数。
解法：还是DP，主要是如何处理'*'。
Java:
public int numDecodings(String s) {
/* initial conditions */
long[] dp = new long[s.length()+1];
dp[0] = 1;
if(s.charAt(0) == '0'){
return 0;
}
dp[1] = (s.charAt(0) == '*') ? 9 : 1;
/* bottom up method */
for(int i = 2; i <= s.length(); i++){
char first = s.charAt(i-2);
char second = s.charAt(i-1);
// For dp[i-1]
if(second == '*'){
dp[i] += 9*dp[i-1];
}else if(second > '0'){
dp[i] += dp[i-1];
}
// For dp[i-2]
if(first == '*'){
if(second == '*'){
dp[i] += 15*dp[i-2];
}else if(second <= '6'){
dp[i] += 2*dp[i-2];
}else{
dp[i] += dp[i-2];
}
}else if(first == '1' || first == '2'){
if(second == '*'){
if(first == '1'){
dp[i] += 9*dp[i-2];
}else{ // first == '2'
dp[i] += 6*dp[i-2];
}
}else if( ((first-'0')*10 + (second-'0')) <= 26 ){
dp[i] += dp[i-2];
}
}
dp[i] %= 1000000007;
}
/* Return */
return (int)dp[s.length()];
}
Python:
class Solution(object):
def numDecodings(self, s):
"""
:type s: str
:rtype: int
"""
if len(s) == 0 or s[0] == '0':
return 0
dp = [0] * (len(s) + 1)
dp[0] = 1
dp[1] = 9 if s[0] == '*' else 1
for i in xrange(2, len(dp)):
first = s[i-2]
second = s[i-1]
# for dp[i-1]
if second == '*':
dp[i] = dp[i-1] * 9
elif second != '0':
dp[i] = dp[i-1]
# for dp[i-2]
if first == '*':
if second == '*':
dp[i] += 15 * dp[i-2]
elif second <= '6':
dp[i] += 2 * dp[i-2]
else:
dp[i] += dp[i-2]
elif first == '1' or first == '2':
if second == '*':
if first == '1':
dp[i] += 9 * dp[i-2]
else:
dp[i] += 6 * dp[i-2]
elif first == '1' or (first == '2' and second <= '6'):
dp[i] += dp[i-2]
dp[i] %= 1000000007
return dp[-1]
Python:
class Solution(object):
def numDecodings(self, s):
"""
:type s: str
:rtype: int
"""
M, W = 1000000007, 3
dp = [0] * W
dp[0] = 1
dp[1] = 9 if s[0] == '*' else dp[0] if s[0] != '0' else 0
for i in xrange(1, len(s)):
if s[i] == '*':
dp[(i + 1) % W] = 9 * dp[i % W]
if s[i - 1] == '1':
dp[(i + 1) % W] = (dp[(i + 1) % W] + 9 * dp[(i - 1) % W]) % M
elif s[i - 1] == '2':
dp[(i + 1) % W] = (dp[(i + 1) % W] + 6 * dp[(i - 1) % W]) % M
elif s[i - 1] == '*':
dp[(i + 1) % W] = (dp[(i + 1) % W] + 15 * dp[(i - 1) % W]) % M
else:
dp[(i + 1) % W] = dp[i % W] if s[i] != '0' else 0
if s[i - 1] == '1':
dp[(i + 1) % W] = (dp[(i + 1) % W] + dp[(i - 1) % W]) % M
elif s[i - 1] == '2' and s[i] <= '6':
dp[(i + 1) % W] = (dp[(i + 1) % W] + dp[(i - 1) % W]) % M
elif s[i - 1] == '*':
dp[(i + 1) % W] = (dp[(i + 1) % W] + (2 if s[i] <= '6' else 1) * dp[(i - 1) % W]) % M
return dp[len(s) % W]
C++：
class Solution {
public:
int numDecodings(string s) {
int n = s.size(), M = 1e9 + 7;
vector<long> dp(n + 1, 0);
dp[0] = 1;
if (s[0] == '0') return 0;
dp[1] = (s[0] == '*') ? 9 : 1;
for (int i = 2; i <= n; ++i) {
if (s[i - 1] == '0') {
if (s[i - 2] == '1' || s[i - 2] == '2') {
dp[i] += dp[i - 2];
} else if (s[i - 2] == '*') {
dp[i] += 2 * dp[i - 2];
} else {
return 0;
}
} else if (s[i - 1] >= '1' && s[i - 1] <= '9') {
dp[i] += dp[i - 1];
if (s[i - 2] == '1' || (s[i - 2] == '2' && s[i - 1] <= '6')) {
dp[i] += dp[i - 2];
} else if (s[i - 2] == '*') {
dp[i] += (s[i - 1] <= '6') ? (2 * dp[i - 2]) : dp[i - 2];
}
} else { // s[i - 1] == '*'
dp[i] += 9 * dp[i - 1];
if (s[i - 2] == '1') dp[i] += 9 * dp[i - 2];
else if (s[i - 2] == '2') dp[i] += 6 * dp[i - 2];
else if (s[i - 2] == '*') dp[i] += 15 * dp[i - 2];
}
dp[i] %= M;
}
return dp[n];
}
};
C++:
class Solution {
public:
int numDecodings(string s) {
long e0 = 1, e1 = 0, e2 = 0, f0, f1, f2, M = 1e9 + 7;
for (char c : s) {
if (c == '*') {
f0 = 9 * e0 + 9 * e1 + 6 * e2;
f1 = e0;
f2 = e0;
} else {
f0 = (c > '0') * e0 + e1 + (c <= '6') * e2;
f1 = (c == '1') * e0;
f2 = (c == '2') * e0;
}
e0 = f0 % M;
e1 = f1;
e2 = f2;
}
return e0;
}
};
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[LeetCode] 91. Decode Ways 解码方法