标签：
A message containing letters from A-Z is being encoded to numbers using the following
mapping:
‘A‘ -> 1 ‘B‘ -> 2 ... ‘Z‘ -> 26
Given an encoded message containing digits, determine the total number of ways to decode it.
For example,
Given encoded message "12", it could be decoded as "AB" (1
2) or "L" (12).
The number of ways decoding "12" is 2.
暴力解法，超时
public int NumDecodings(string s)
{
if (s.Length == 0)
return 0;
if (s.Length ==1&&s[0]!='0')
return 1;
if (s[0] == '0')
return 0;
int tmp = s.Length == 2 ? 1 : 0;
if (s[0] == '1' || (s[0] == '2' && s[1] < '7'))
return NumDecodings(s.Substring(1)) + NumDecodings(s.Substring(2)) + tmp;
return NumDecodings(s.Substring(1));
}如果当前数字为0，则前面数字必为1或者2，否则无法进行编码转换，此时只能和前面的1或者2连在一起进行编码，因此res[i] = res[i-2]；
如果前面数字和前面数字的组合可以进行2种编码（分开或者和在一起），则res[i] = res[i-1] + res[i-2]；
否则，当前数字必须单独进行编码转换，res[i] = res[i-1]。
public int NumDecodings(string s)
{
if (s.Length == 0 || s[0] == '0')
return 0;
int[] res = new int[s.Length + 1];
res[0] = res[1] = 1;
for (int i = 2; i <= s.Length; i++)
{
if (s[i-1]=='0')
{
if (s[i - 2] == '1'|| s[i - 2] == '2')
{
res[i] = res[i - 2];
}
else
{
return 0;
}
}
else
{
res[i] = res[i - 1];
if (s[i-2]!='0')
{
int val = (s[i - 2] - '0') * 10 + s[i - 1] - '0';
if (0 < val && val < 27)
res[i] += res[i - 2];
}
}
}
return res[s.Length];
}再改进，倒序
public int NumDecodings(string s)
{
int n = s.Length;
if (n == 0) return 0;
int[] res = new int[n + 1];
res[n] = 1;
res[n - 1] = s[n - 1] == '0' ? 0 : 1;
for (int i = n - 2; i >= 0; i--)
if (s[i] != '0')
res[i] = (Convert.ToInt16((s.Substring(i,2))) < 27) ? res[i + 1] + res[i + 2] : res[i + 1];
return res[0];
}标签：
原文地址：http://blog.csdn.net/zmq570235977/article/details/51330936
