91.Decode-Ways.md

91. Decode Ways

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题目地址

https://leetcode.com/problems/decode-ways/

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 a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:
Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Example 2:
Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

代码

Approach 1: Recursive with memoization

class Solution {

  HashMap<Integer, Integer> memo = new HashMap<>();

  public int numDecodings(String s) {
    if (s == null || s.length() == 0) {
      	return 0;
    }

    return recursiveWithMemo(0, s);
  }

  private int recursiveWithMemo(int index, String str) {
    // index + 1 or index + 2
    if (index == str.length() || index == str.length() - 1) {
      return 1;
    }
    // Ways to decode a string of size 1 is 1. Unless the string is '0'.
    if (str.charAt(index) == '0') {
      return 0;
    }

    if (memo.containsKey(index)) {
      return memo.get(index);
    }

    // index + 1
    int ans = recursiveWithMemo(index + 1, str);
    // index + 2
    if (Integer.parseInt(str.substring(index, index + 2)) <= 26)
    {
      ans += recursiveWithMemo(index + 2, str);
    } 

    memo.put(index, ans);

    return ans;
  }
}

Approach #2 Dynamic Programming

Complexity Analysis

• Time Complexity: O(N), where N is length of the string. We iterate the length of dp array which is N+1.
• Space Complexity: O(N). The length of the DP array.

class Solution {
  public int numDecodings(String s) {
    if (s == null || s.length() == 0) return 0;

    int[] dp = new int[s.length() + 1];
    dp[0] = 1;
    dp[1] = s.charAt(0) == '0' ? 0 : 1;
    for (int i = 2; i < dp.length; i++) {
      // one digit
      if (s.charAt(i - 1) != '0') {
        dp[i] += dp[i - 1];
      }
			// two digit
      int twoDigit = Integer.valueOf(s.substring(i - 2, i));
      if (twoDigit >= 10 && twoDigit <= 26) {
        dp[i] += dp[i - 2];
      }
    }

    return dp[s.length()];
  }
}