Dynamic programming (integer division) --- letter string division of integers

Letter string division of integers

91. Decode Ways (Medium)

Subject description:

A message containing letters from A-Z is being encoded to numbers using the following mapping:

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

Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

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

Example 2:

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

Analysis of ideas:

  Dynamic Programming Solutions, for each number, it can only be decoded in two ways, namely itself and its former or ten-digit numbers. We define dp [i]: Method Number before decoding the i-th digit. Then if s [i] between the [1-9], dp [i] shall add dp [i-1], between if s [i-1] + s [i] in [10, 26] , then dp [i] shall add dp [i-2], the total number thus obtained is dp [i] is the number of the decoding method. It is to be noted that s [i] is 0, this is the default if s [i] is 0, s [i-1] + s [i] between affirmative [10-26] (or decoding received no ), if s are all 0, we have to exclude this case.

Code:

public int numDecodings(String s){
    if(s==null||s.length()==0)
        return 0;
    int n=s.length();
    int []dp=new int[n+1];
    dp[0]=1;
    dp[1]=s.charAt(0)=='0'?0:1; //第一个字符为0，则无法解码，否则解码方法数为1
    for(int i=2;i<=n;i++){
        int one=Integer.valueOf(s.subString(i-1,i)); //以其本身解码
        if(one!=0)
            dp[i]+=dp[i-1];
        if(s.charAt(i-2)=='0') //如果当前字符的前一个字符为0，那么当前字符不能和前一个字符组成十位数进行解码
            continue;
        int two=Integer.valueOf(s.subString(i-2,i));
        if(two<=26)
            dp[i]=dp[i]+dp[i-2];
    }
    return dp[n];
}