LeetCode Decode Ways

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Description:

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.

Solution:

本来考虑的是每次遇到11-19和21-26的数字就乘以2，但是后来发现如果遇到了1121的情况就很难用乘以2的方式表达。以及遇到10和20需要固定住这两位数字。

换一个方式来，用状态转移来表达。

dp[i]表示到了第i位一共有多少种情况，有这么几种转移的方式：

1. nums[i]在11-19和21-26之间，那么dp[i] = dp[i-1] + dp[i-2]

2. nums[i]是10或者20，dp[i] = dp[i-2]

3. 其余情况dp[i] = dp[i-1]

import java.util.*;public class Solution {public int numDecodings(String s) {if (s.equals(""))return 0;int len = s.length();// 包含0，00，30，40到90的情况for (int i = 0; i < len; i++) {if (s.charAt(i) == '0') {if (i == 0 || s.charAt(i - 1) == '0' || s.charAt(i - 1) >= '3')return 0;}}// 放在for循环的后面，是为了防止出现“0”的情况if (s.length() == 1)return 1;int[] dp = new int[len];dp[0] = 1;char ch1, ch2;int temp;ch1 = s.charAt(0);ch2 = s.charAt(1);temp = (ch1 - '0') * 10 + (ch2 - '0');if (canDivide(temp))dp[1] = 2;elsedp[1] = 1;for (int i = 2; i < len; i++) {ch1 = s.charAt(i - 1);ch2 = s.charAt(i);temp = (ch1 - '0') * 10 + (ch2 - '0');if (canDivide(temp))dp[i] = dp[i - 1] + dp[i - 2];else if (temp == 10 || temp == 20)dp[i] = dp[i - 2];elsedp[i] = dp[i - 1];}return dp[len - 1];}boolean canDivide(int t) {if (t >= 11 && t <= 19)return true;else if (t >= 21 && t <= 26)return true;return false;}}

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