【LeetCode】129.Decode Ways

最新推荐文章于 2020-02-27 10:19:39 发布

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最新推荐文章于 2020-02-27 10:19:39 发布

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本文探讨了LeetCode上一道中等难度的问题——解码方式。通过动态规划的方法，我们解析了如何计算给定数字字符串的解码方式总数。文章详细介绍了算法思路，并提供了完整的C++代码实现。

题目描述(Medium)

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.

题目链接

https://leetcode.com/problems/decode-ways/description/

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).

算法分析

动态规划，cur = 当前为两位数的解法数prev + 当前为一位数的解法数cur，加上限制条件即可。

提交代码：

class Solution {
public:
    int numDecodings(string s) {
        if(s.empty() || s[0] == '0') return 0;
        const int n = s.size();
        int prev = 0, cur = 1;
        
        for (int i = 0; i < n; ++i) {
            // 当前为1位数解法不可行
            if (s[i] == '0') cur = 0;
            
            // 当前为2位数解法不可行
            if (i < 1 || !(s[i - 1] == '1' || 
                           s[i - 1] == '2' && s[i] <= '6'))
                prev = 0;
            
            int temp = cur;
            cur += prev;
            prev = temp;
        }
        
        return cur;
    }
};