Leetcode: Decode Ways

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最新推荐文章于 2025-05-16 12:14:55 发布

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分类专栏： Leetcode 文章标签： leetcode

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本文介绍了一种使用动态规划解决消息解码问题的方法。对于由A到Z字母组成的编码消息，每种字母对应一个数字（A-1，B-2...Z-26），文章详细解释了如何计算出所有可能的解码方式数量。

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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.

Dynamic programming: number of ways decoding message before position i is f [ i ]. If the number at i together with the number at i - 1 can be decoded to a valid letter, then f [ i ] = f [i - 1] + f [i - 2]. Else, f [ i ] = f [ i - 1]. Need to pay attention to 0 cases.

public class Solution {
    public int numDecodings(String s) {
        if (s == null || s.length() == 0 || s.charAt(0) == '0') {
            return 0;
        }
        
        int[] num = new int[s.length()];
        num[0] = 1;
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i) == '0') {
                if (s.charAt(i - 1) - '0' < 1 || s.charAt(i - 1) - '0' > 2) {
                    return 0;
                } else {
                    num[i] = (i == 1) ? 1 : num[i - 2];
                }
            } else {
                if (s.charAt(i - 1) - '0' > 0 && (s.charAt(i - 1) - '0' == 1 || 
                    (s.charAt(i - 1) - '0' == 2 && s.charAt(i) - '0' <= 6))) {
                        num[i] = (i == 1) ? 2 : num[i - 1] + num[i - 2];
                } else {
                    num[i] = num[i - 1];
                }
            }
        }
        
        return num[s.length() - 1];
    }
}