Leetcode 91.Decode Ways (decoding mode) thinking and method of solving problems

A message containing letters from was A-Z 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 " , it could be decoded as " AB "   (1 2) Or " L "   ().

The number of ways decoding is "12" 2.

Ideas: The subject with recursion is not realized, it may be that the recursion is not written well, and finally in the online reference to write the dynamic programming code.

The idea is to judge whether the current value is 0, not 0 f[i] = f[i] + f[i-1];

Then the current value and the previous character can be composed of 26 or less of the loss, can this with f[i] = F[i] + f[i-2].

The specific code is as follows:

public class Solution {public    int numdecodings (String s) {    //dynamic plan tag    int[] f = new int[s.length ()];    Char[] C = S.tochararray ();    Boundary condition    if (c.length = = 0) {    return 0;    }    First element    f[0] = c[0] > ' 0 '? 1:0;        if (c.length = = 1) {    return f[0];    }    F[1] value is key, write bad, there will be various errors    int k = c[0] > ' 0 ' && c[1] > ' 0 '? 1:0;    F[1] = k + (c[0] = = ' 1 ' | | c[0] = = ' 2 ' && c[1] <= ' 6 '? 1:0);        Iterate backwards    for (int i = 2; i < c.length; i++) {    if (C[i] > ' 0 ') {//first element greater than 0, add    f[i] + = f[i-1];    }    //Between 10-26 add two letters to form a case    if (c[i-1] = = ' 1 ' | | (C[i-1] = = ' 2 ' && c[i] <= ' 6 ')) {    F[i] + = F[i-2];}    }        return f[c.length-1];}    }

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Leetcode 91.Decode Ways (decoding mode) thinking and method of solving problems