91. Decode Ways

问题：

给定由0～9构成的编码字符串。

编码前，

• 'A' -> "1"
• 'B' -> "2"
• ...
• 'Z' -> "26"

 

按照以上方法进行编码。

求编码前的字符串组合的可能数。

Example 1:
Input: s = "12"
Output: 2
Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).

Example 2:
Input: s = "226"
Output: 3
Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Example 3:
Input: s = "0"
Output: 0
Explanation: There is no character that is mapped to a number starting with 0.
The only valid mappings with 0 are 'J' -> "10" and 'T' -> "20", neither of which start with 0.
Hence, there are no valid ways to decode this since all digits need to be mapped.

Example 4:
Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").
 
Constraints:
1 <= s.length <= 100
s contains only digits and may contain leading zero(s).

　　

解法：DP(动态规划)

1.确定【状态】：

• 字符串s的index：i

2.确定【选择】：

dp[i]：SUM {

• 两个数字->一个字母
  • dp[i-2]
    • 条件：s[i-1]==2 && s[i]<=6
    • or       s[i-1]==1
• 一个数字->一个字母
  • dp[i-1]
    • 条件：s[i]!=0
    • if s[i]==0 then return 0.
• }

3. dp[i]的含义：

• 字符串s[0~i]  能够构成编码的可能数。 

4. 状态转移：

dp[i]= SUM {

• if((s[i-1]==2 && s[i]<=6) || s[i-1]==1) dp[i-2]
  • others: 0
• if(s[i]!=0) dp[i-1]
  • others: 0
• }

5. base case：

• dp[1]= (s[0]!=0)? 1 : 0;
• dp[0]=1;//如 "12"->直接编码为"H"，dp[2]+=dp[0]，能够加出一个1。

 

代码参考：

class Solution {
public:
    //dp[i]: the number of ways to decode s[0~i]
    //opt: case_1+case_2
    //case_1: dp[i-1]
    //case_2: if((s[i-1]==2 && s[i]<=6) || s[i-1]==1) dp[i-2]
    //base:dp[0]=0  dp[1]=1
    int numDecodings(string s) {
        int pre_0=1, pre_1=1, res=pre_1;
        int len = s.length();
        if(s[0]<='0') return 0;
        for(int i=1; i<len; i++) {
            if(s[i]>'0') res = pre_1;
            else res = 0;
            //cout<<res<<endl;
            if((s[i-1]=='2' && s[i]<='6') || s[i-1]=='1') res += pre_0;
            pre_0 = pre_1;
            pre_1 = res;
        }
        return res;
    }
};