leetcode — decode-ways

/**
 * Source : https://oj.leetcode.com/problems/decode-ways/
 *
 *
 * 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.
 */
public class DecodeWays {

    /**
     * 找出有多少种解码方式
     *
     * 使用递归，但是复杂度较高，可以考虑使用DP
     *
     * 当XY > 26的时候 dp[i+1] = dp[i]
     * XY <= 26 的时候 dp[i+1] = dp[i] + dp[i-1]
     *
     *
     * 临界条件：
     * X = 0,dp[i+1] = dp[i]
     * Y = 0,dp[i+1] = dp[i-1]
     *
     * @param digits
     */
    public int findWays (String digits) {
        if (digits == null || digits.length() == 0 || digits.charAt(0) < '1' || digits.charAt(0) > '9') {
            return 0;
        }
        int[] dp = new int[digits.length() + 1];
        dp[0] = dp[1] = 1;
        for (int i = 1; i < digits.length(); i++) {
            if (digits.charAt(i) > '9' || digits.charAt(i) < '1') {
                return 0;
            }
            int x = digits.charAt(i-1) - '0';
            int y = digits.charAt(i) - '0';
            int xy = x * 10 + y;
            if (xy > 9 && xy <= 26) {
                dp[i+1] = dp[i] + dp[i-1];
            } else if (y != 0) {
                dp[i+1] = dp[i];
            }
            if (dp[i+1] == 0) {
                return 0;
            }
        }
        return dp[dp.length-1];

    }

    public static void main(String[] args) {
        DecodeWays decodeWays = new DecodeWays();
        System.out.println(decodeWays.findWays(""));
        System.out.println(decodeWays.findWays("1"));
        System.out.println(decodeWays.findWays("12"));
        System.out.println(decodeWays.findWays("32"));
        System.out.println(decodeWays.findWays("10"));
        System.out.println(decodeWays.findWays("00"));
        System.out.println(decodeWays.findWays("09"));
    }
}