91. Decode Ways

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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.

这道题需要用到DP。建立数组存储每一位之后的组合数。从后向前遍历，如果这一位跟后面一位组成的数小于26，说明可以有新的组合，count[i] = count[i+1] + count[i+2]，如果大于26，说明还是按部就班，count[i] = count[i+1]。代码如下：

public class Solution {    public int numDecodings(String s) {        if (s == null || s.length() == 0) {            return 0;        }        int len = s.length();        char[] chs = s.toCharArray();        int[] count = new int[len + 1];        count[len] = 1;        count[len - 1] = chs[len - 1] == '0'? 0 : 1;        for (int i = len - 2; i >= 0; i --) {            if (chs[i] == '0') {                continue;            }            int num = (chs[i] - '0') * 10 + (chs[i + 1] - '0');            if (num <= 26) {                count[i] = count[i + 1] + count[i + 2];            } else {                count[i] = count[i + 1];            }        }        return count[0];    }}

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