Decode Ways

描述

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.

分析

跟第???节Climbing Stairs很类似，不过多加一些判断逻辑。

代码

Python

// Decode Ways
// 动规，时间复杂度O(n)，空间复杂度O(1)
public class Solution {
    public int numDecodings(String s) {
        if (s.isEmpty() || s.charAt(0) == '0') return 0;

        int prev = 0;
        int cur = 1;
        // 长度为n的字符串，有 n+1个阶梯
        for (int i = 1; i <= s.length(); ++i) {
            if (s.charAt(i-1) == '0') cur = 0;

            if (i < 2 || !(s.charAt(i - 2) == '1' ||
                    (s.charAt(i - 2) == '2' && s.charAt(i - 1) <= '6')))
                prev = 0;

            int tmp = cur;
            cur = prev + cur;
            prev = tmp;
        }
        return cur;
    }
}

Java

// Decode Ways
// 动规，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    int numDecodings(const string &s) {
        if (s.empty() || s[0] == '0') return 0;

        int prev = 0;
        int cur = 1;
        // 长度为n的字符串，有 n+1个阶梯
        for (size_t i = 1; i <= s.size(); ++i) {
            if (s[i-1] == '0') cur = 0;

            if (i < 2 || !(s[i - 2] == '1' ||
                     (s[i - 2] == '2' && s[i - 1] <= '6')))
                prev = 0;

            int tmp = cur;
            cur = prev + cur;
            prev = tmp;
        }
        return cur;
    }
};

C++

# Decode Ways
# Dynamic programming, time complexity O(n), space complexity O(1)
class Solution:
    def numDecodings(self, s: str) -> int:
        if not s or s[0] == '0':
            return 0

        prev = 0
        cur = 1
        # For a string of length n, there are n+1 steps
        for i in range(1, len(s) + 1):
            if s[i-1] == '0':
                cur = 0

            if i < 2 or not (s[i-2] == '1' or
                    (s[i-2] == '2' and s[i-1] <= '6')):
                prev = 0

            tmp = cur
            cur = prev + cur
            prev = tmp

        return cur

相关题目

• Climbing Stairs