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.

分析

斐波那契数列和台阶问题的变形，只不过加了一个条件判断。

代码

    public static int numDecodings(String s) {

        if (s == null || s.length() == 0) {
            return 0;
        }

        int n = s.length();
        char[] c = s.toCharArray();

        // 对于台阶，需要前两步的值，所以数组最小是3
        int[] step = new int[Math.max(n + 1, 3)];

        step[0] = 1;
        step[1] = 0;

        // 第一个字符不是0，则第一步初始为1
        if (c[0] != '0') {
            step[1] = 1;
        }

        // step[i] = step[i - 1] + step[i - 2];
        // 只不过加step[i - 2]时，需要对c[i - 2]和c[i - 1]判断，组合是否<=26
        for (int i = 2; i <= n; i++) {

            step[i] = 0;

            if (c[i - 1] != '0') {
                step[i] += step[i - 1];
            }

            if (c[i - 2] != '0') {
                if ((c[i - 2] - '0') * 10 + (c[i - 1] - '0') <= 26) {
                    step[i] += step[i - 2];
                }
            }
        }

        return step[n];
    }