Decode Ways

题目描述

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.

题目解答

题目分析

从头到尾扫这个String，比如我们想知道到，从第一位到dp[i]这一位组成的String，有多少种解码组合，那么有两种情况

第一：如果dp[i]所对应的的单个字符可以解码，那么dp[i]就包括前dp[i-1]位所积累的组合数 dp[i] = dp[i-1]

第二：如果不仅dp[i]所对应的的单个字符可以解码，dp[i-1] － dp[i]，两个字符组成的也可以解码，那么不仅包括dp[i-1]积累的组合数，也包括dp[i-2]位积累的组合数 dp[i] = dp[i-1] + dp[i-2]

代码实现

public class Solution {
    public int numDecodings(String s) {
        if(s == null || s.length() == 0 || s.charAt(0) == '0')
            return 0;
        int len = s.length();
        int[] dp = new int[len+1];
        dp[0] = 1;
        if(isValid(s.substring(0,1)))
            dp[1] = 1;
        else 
            dp[1] = 0;
        for(int i = 2; i <= len; i++) {
            if(isValid(s.substring(i-1, i)))
                dp[i] = dp[i-1];
            if(isValid(s.substring(i-2, i)))
                dp[i] += dp[i-2];
        }
        return dp[len];
    }
    public boolean isValid(String s) {
        if(s.charAt(0) == '0')
            return false;
        int num = Integer.parseInt(s);
        return num >= 0 && num <= 26;
    }
}