0091. Decode Ways - kumaeki/LeetCode GitHub Wiki

0091. Decode Ways

A message containing letters from A-Z can be encoded into numbers using the following mapping:

  • 'A' -> "1"
  • 'B' -> "2" ...
  • 'Z' -> "26"

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:

  • "AAJF" with the grouping (1 1 10 6)
  • "KJF" with the grouping (11 10 6) Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".

Given a string s containing only digits, return the number of ways to decode it.

The answer is guaranteed to fit in a 32-bit integer.

Example 1:

Input: s = "12"

Output: 2

Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: s = "226"

Output: 3

Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Example 3:

Input: s = "0"

Output: 0

Explanation: There is no character that is mapped to a number starting with 0.
The only valid mappings with 0 are 'J' -> "10" and 'T' -> "20", neither of which start with 0.
Hence, there are no valid ways to decode this since all digits need to be mapped.

Example 4:

Input: s = "06"

Output: 0

Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").

Constraints:

  • 1 <= s.length <= 100
  • s contains only digits and may contain leading zero(s).

解法1

dfs搜索. 但是太慢了.

class Solution {
    public int numDecodings(String s) {
        return helper(s, 0);
    }
    
    private int helper(String s, int index){
        if(index > s.length())
            return 0;
        
        if(index == s.length())
            return 1;
        
        char c = s.charAt(index);
        // 0 不是有效数字
        if(c == '0')
            return 0;
        
        // 0以外的一位数字都是有效的
        int result = helper(s, index + 1);
        
        // 小于27的二位数才是有效数字
        if(c == '1')
            result += helper(s, index + 2);
        else if(c == '2' && index < s.length() - 1 && s.charAt(index + 1) < '7')
            result += helper(s, index + 2);
        
        return result;
            
    }
}

解法2

用memo来储存之前计算过的内容, 避免重复.

class Solution {
    public int numDecodings(String s) {
        if(s==null || s.length()==0)
            return 1;
        if(s.charAt(0) == '0')
            return 0;
        
        int n = s.length(), pre = 1, cur = 1, res = 0;

        for(int i = 1; i < n; i++){
            char ccur = s.charAt(i);
            char cpre = s.charAt(i-1);
            if(ccur != '0'){
                res = cur;
                if(cpre == '1' || (cpre == '2' && ccur < '7'))
                    res += pre;
            }else{
                if(cpre == '0' || cpre > '2')
                    return 0;
                else
                    res = pre;
            }
            pre = cur;
            cur = res;
        }
        return cur;
    }
}
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