标签：turn val win 解析 div ring numbers i++ following
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.
DP:
设置动态数组dp[n+1]。dp[i]表示从1~i的decode ways的个数。
当给的code只有一位数时，判断是不是valid（A~Z），是的话就dp[1] = 1 不是的话就是dp[1] = 0
因为像给的例子12可以有两种可能的解析方法，所以计算dp[i]的时候要判断两种可能性，再累加
public class Solution { public int numDecodings(String s) { if(s == null || s.length() == 0) return 0; int dp[] = new int[s.length() + 1]; dp[0] = 1; if(isValid(s.substring(0,1))) dp[1] = 1; else return 0; for(int i = 2; i <= s.length(); 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[s.length()]; } public boolean isValid(String s){ if( s.charAt(0) == ‘0‘) return false; int temp = Integer.parseInt(s); if(temp > 0 && temp <= 26) return true; return false; } }
标签：turn val win 解析 div ring numbers i++ following
原文地址：http://www.cnblogs.com/joannacode/p/5991669.html
