[LeetCode] 91. Decode Ways 解码方法

Posted 2020-10-25 轻风舞动

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.

思路：

用动态规划Dynamci Programming来解。一位数时不能为0，两位数不能大于26，其十位上的数也不能为0。用哈希表来存或者用两个变量来存。

State: dp[i]

Function: dp[i] = dp[i - 1] (if i - 1 != 0) + dp[i - 2] (if i - 2 == 1 or i - 2 == 2 and i -1 < = 6)

Initialize: dp[0] = 0, dp[1] = 1

Return: dp[n]

Java 1:

public class Solution {
    public int numDecodings(String s) {
        if (s.isEmpty() || (s.length() > 1 && s.charAt(0) == ‘0‘)) return 0;
        int[] dp = new int[s.length() + 1];
        dp[0] = 1;
        for (int i = 1; i < dp.length; ++i) {
            dp[i] = (s.charAt(i - 1) == ‘0‘) ? 0 : dp[i - 1];
            if (i > 1 && (s.charAt(i - 2) == ‘1‘ || (s.charAt(i - 2) == ‘2‘ && s.charAt(i - 1) <= ‘6‘))) {
                dp[i] += dp[i - 2];
            }
        }
        return dp[s.length()];
    }
}

Java 2:

public class Solution {
    public int numDecodings(String s) {
        if (s.isEmpty() || (s.length() > 1 && s.charAt(0) == ‘0‘)) return 0;
        int prev = 1, prev_prev = 0;
        for (int i = 0; i < s.length(); i++) {
            int cur = 0;
            if (s.charAt(i) != ‘0‘) {
                cur = prev;
            }
            if (i > 0 && (s.charAt(i - 1) == ‘1‘ || (s.charAt(i - 1) == ‘2‘ && s.charAt(i - 1) <= ‘6‘))) {
                cur += prev_prev;
            }
            prev_prev = prev;
            prev = cur;
        }
        return prev;
    }
}

　　

Python:

class Solution(object):
    def numDecodings(self, s):
        if len(s) == 0 or s[0] == ‘0‘:
            return 0
        prev, prev_prev = 1, 0
        for i in xrange(len(s)):
            cur = 0
            if s[i] != ‘0‘:
                cur = prev
            if i > 0 and (s[i - 1] == ‘1‘ or (s[i - 1] == ‘2‘ and s[i] <= ‘6‘)):
                cur += prev_prev
            prev, prev_prev = cur, prev
        return prev