Monday, April 6, 2015

LeetCode [91] Decode Ways

91. Decode Ways
Medium

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.

The answer is guaranteed to fit in a 32-bit integer.

 

Example 1:

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

Example 2:

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

Example 3:

Input: s = "0"
Output: 0
Explanation: There is no character that is mapped to a number starting with '0'. We cannot ignore a zero when we face it while decoding. So, each '0' should be part of "10" --> 'J' or "20" --> 'T'.

Example 4:

Input: s = "1"
Output: 1

 

Constraints:

  • 1 <= s.length <= 100
  • s contains only digits and may contain leading zero(s).
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class Solution
{
public:
    int numDecodings(string s)
    {
        int n = s.size();
        vector<int> dp(n, 0);

        for (int i = 0; i < n; ++i)
        {
            int v = 0;
            if (i > 0)
                v = stoi(s.substr(i - 1, 2));

            if (s[i] == '0')
            {
                if (v == 10 || v == 20)
                    dp[i] = i > 1 ? dp[i - 2] : 1;
            }
            else
            {
                dp[i] = i > 0 ? dp[i - 1] : 1;
                if (v >= 10 && v <= 26)
                {
                    dp[i] += i > 1 ? dp[i - 2] : 1;
                }
            }
        }
        return dp[n - 1];
    }
};

class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        if(n==0) return n;

        int dp[10000] = {0};
        dp[n] = 1;
        dp[n-1] = s[n-1]=='0'?0:1;

        for(int i=n-2; i>=0; --i){
            if(s[i]=='0') continue;
            int v = stoi(s.substr(i,2));
            dp[i] = v<=26?dp[i+1]+dp[i+2]:dp[i+1];
        }
        return dp[0];
    }
};

#define N 10000
class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        if(!n) return 0;

        int dp[N] = {0};
        dp[0] = 1;
        for(int i=1; i<=n; ++i){
            if(s[i-1]!='0') 
                dp[i] = dp[i-1];
            if(i-2>=0)
                if(s[i-2]=='1'||(s[i-2]=='2'&&s[i-1]>='0'&&s[i-1]<='6')) dp[i] += dp[i-2];
        }

        return dp[n];
    }
};

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