字符串分解问题

[LeetCode 91] Decode Ways

题目

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.

测试案例

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

Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

思路

本题的主要细节是；对当前一位进行解码时，当前位不能为0对当前两位进行解码时，当前两位数的值应大于9且小于27.

代码如下

class Solution {
    public int numDecodings(String s) {
        int n = s.length(), pre, cur;
        if(n == 0){
            return 0;
        }        
        int[] nums = new int[n + 1];
        nums[0] = 1;
        nums[1] = s.charAt(0) == '0' ? 0 : 1;
        for(int i = 2; i <= n; i++){
            pre = s.charAt(i - 2) - '0';
            cur = s.charAt(i - 1) - '0';
            if(cur > 0){
                nums[i] += nums[i - 1];
            }            
            int value = pre * 10 + cur;
            if(value > 9 && value < 27){                
                nums[i] += nums[i - 2];
            }
        }
        return nums[n];
    }
}

[LeetCode 639] Decode Ways II

题目

A message containing letters from A-Z is being encoded to numbers using the following mapping way:

'A' -> 1
'B' -> 2
...
'Z' -> 26
Beyond that, now the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9.
Given the encoded message containing digits and the character '*', return the total number of ways to decode it.
Also, since the answer may be very large, you should return the output mod 10^9 + 7.

Note:
The length of the input string will fit in range [1, 10^5].
The input string will only contain the character '*' and digits '0' - '9'.

测试案例

Input: "*"
Output: 9
Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".

Input: "1*"
Output: 9 + 9 = 18

思路

1. 对当前位进行解码时，当前位分：0，1~9，*三种情况进行讨论。
2. 对当前两位进行解码时，当前两位分：
   • 全为*：有 15nums[i - 2] 种情况
   • *非*：个位介于0~6之间时，有 2nums[i - 2] 种情况；个位大于6时，有 nums[i - 2] 种情况
   • 十位为非*，个位为*：十位分 0或3~9，1，2 三种情况考虑。
   • 均非*：两位数要大于9，且小于的等于26.

代码如下

class Solution {    
    public int numDecodings(String s) {
        final int MODULE = 1_000_000_000 + 7;
        int n = s.length(), pre, cur, value;
        if(n == 0){
            return 0;
        }        
        long[] nums = new long[n + 1];
        nums[0] = 1;
        char ch;
        if((ch = s.charAt(0)) == '0'){
            nums[1] = 0;
        }
        else if(ch == '*'){
            nums[1] = 9;
        }
        else{
            nums[1] = 1;
        }        
        for(int i = 2; i <= n; i++){
            pre = s.charAt(i - 2);
            cur = s.charAt(i - 1);
            long temp = 0;
            if(cur >= '1' && ch <= '9'){
                temp = nums[i - 1];
            }                   
            else if(cur == '*'){
                temp = 9 * nums[i - 1];
            }
            nums[i] = (nums[i] + temp) % MODULE;
            temp = 0;
            if(pre != '*' && cur != '*'){                
                value = (pre - '0') * 10 + cur - '0';
                if(value > 9 && value < 27){                
                    temp = nums[i - 2];
                }
            }            
            else if(cur == '*' && pre != '*'){
                if(pre == '1'){
                    temp = 9 * nums[i - 2];  
                }         
                else if(pre == '2'){
                    temp = 6 * nums[i - 2];  
                }
            }
            else if(pre == '*' && cur != '*'){
                if(cur < '7'){
                    temp = 2 * nums[i - 2];  
                }
                else{
                    temp = nums[i - 2];  
                }
            }
            else{
                temp = 15 * nums[i - 2];  
            }
            nums[i] = (nums[i] + temp) % MODULE;
        }
        return (int)nums[n];
    }
}