91. Decode Ways

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#Decode Ways

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本文介绍了一种用于解码数字字符串的算法，该算法基于字母到数字的映射规则（例如 'A'->1, 'B'->2, ... 'Z'->26），通过递推公式 f(n)=f(n-1)+f(n-2) 来计算解码方式的数量。文章还提供了一个Java实现案例。

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.

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Solution:

Tips:

f(n) = f(n - 1) + f(n - 2) for character our of range 'A' to 'Z' needs to specific dealing.

Java Code:

public class Solution {
    public int numDecodings(String s) {
        if (s == null || s.length() < 1 || s.charAt(0) == '0') {
            return 0;
        }
        
        int p = 1;
        int q = 1;
        int result = 1;
        for (int i = 1; i < s.length(); i++) {
            char cc = s.charAt(i);
            char pc = s.charAt(i - 1);
            
            if ((pc > '2' || pc == '0') && cc == '0') {
                return 0;
            }
        
            if (cc == '0') {
                result = p;
            } else if (pc == '1' || pc == '2' && cc <= '6') {
                result = p + q;
            } else {
                // pc == '0' || pc == '2' && cc == 7 8 9 || pc >= 3
                result = q;
            }
            p = q;
            q = result;
        }
        
        return result;
    }
}