Decode Ways

问题描述

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

‘A’ -> 1
‘B’ -> 2
…
‘Z’ -> 26

Given a non-empty string containing only digits, determine the total number of ways to decode it.

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).

解题思路

题意：给出数字型字符串，按照 ‘A’ -> 1, …‘Z’ -> 26的形式编码成字符串只包含A-Z的，可以编码成多少种。对于这类问题想到的就是动态规划，想要知道长度为n的解，就必须要知道长度为n-1的解。在此过程中主要考虑字符串中存在‘0’，连续的两个数字构成的两位数大于26的情况。

代码

public int numDecodings(String s) {
	  int n = s.length();
      if(n == 0 ) return 0;
      char[] c = s.toCharArray();
      int[] dp = new int[n];
      // 边界
      if( c[0] == '0') return 0;
      else dp[0] = 1;
      if(n >= 2) {
    	  int m = ((c[0]-'0')*10 + c[1]-'0');
    	  if(c[1] == '0') {
    		  if(m > 26) return 0;
    		  else dp[1] = 1;
    	  }else {
    		  dp[1] = (m > 26 ? 1 : 2);
    	  }
      }
      for(int i = 2; i < n; i++){
          // 连续为零 直接返回
    	  if(c[i] == '0' && c[i-1] == '0') return 0;
    	  int m = ((c[i-1]-'0')*10 + c[i]-'0');
    	  if(c[i - 1] == '0') {
    		  dp[i] = dp[i-1];
    	  }else if(c[i] == '0'){
    		  if(m > 26) return 0;
    		  else dp[i] = dp[i-2];
    	  }else {
    		  dp[i] = dp[i-1] + (m > 26 ? 0 : dp[i-2]);
    	  }
      }
      return dp[n-1];
    }