标签：map data- wing show follow cte ping int final
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 10^9 + 7.
* and digits 0 - 9.
public class Solution {
/**
* @param s: a message being encoded
* @return: an integer
*/
public int numDecodings(String s) {
if (s == null || s.length() == 0) {
return 0;
}
final int mod = 1000000007;
int n = s.length();
int[] f = new int[n + 1];
f[0] = 1;
for (int i = 1; i <= n; i++) {
f[i] = 0;
if (s.charAt(i - 1) == ‘*‘) {
f[i] = (int)((f[i] + 9L * f[i - 1]) % mod);
if (i >= 2) {
if (s.charAt(i - 2) == ‘*‘) {
f[i] = (int)((f[i] + 15L * f[i - 2]) % mod);
}
else if (s.charAt(i - 2) == ‘1‘) {
f[i] = (int)((f[i] + 9L * f[i - 2]) % mod);
}
else if (s.charAt(i - 2) == ‘2‘) {
f[i] = (int)((f[i] + 6L * f[i - 2]) % mod);
}
}
}
else {
if (s.charAt(i - 1) != ‘0‘) {
f[i] = (f[i] + f[i - 1]) % mod;
}
if (i >= 2) {
if (s.charAt(i - 2) == ‘*‘){
if (s.charAt(i - 1) <= ‘6‘) {
f[i] = (int)((f[i] + 2L * f[i - 2]) % mod);
}
else {
f[i] = (f[i] + f[i - 2]) % mod;
}
}
else {
int twoDigits = (s.charAt(i - 2) - ‘0‘) * 10 + s.charAt(i - 1) - ‘0‘;
if (twoDigits >= 10 && twoDigits <= 26) {
f[i] = (f[i] + f[i - 2]) % mod;
}
}
}
}
}
return f[n];
}
}
标签：map data- wing show follow cte ping int final
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078271.html
