[LeetCode] 91. Decode Ways 解码方法

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).

解法1: 递归, Time: O(2^n)

解法2:动态规划Dynamic Programming。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],代表i之前的数字的解法数量。注意index:dp长度比数组多1,所以 s[i-1]是当前数字,dp[i]是当前数字的解法数。

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

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

Return: dp[n]

Java: Recursive

 int numDecodings(string s) {
        return s.empty() ? 0: numDecodings(0,s);    
    }
    int numDecodings(int p, string& s) {
        int n = s.size();
        if(p == n) return 1;
        if(s[p] == '0') return 0;
        int res = numDecodings(p+1,s);
        if( p < n-1 && (s[p]=='1'|| (s[p]=='2'&& s[p+1]<'7'))) res += numDecodings(p+2,s);
        return res;
    }  

Java: Memoization O(n)

int numDecodings(string s) {
        int n = s.size();
        vector<int> mem(n+1,-1);
        mem[n]=1;
        return s.empty()? 0 : num(0,s,mem);   
    }
    int num(int i, string &s, vector<int> &mem) {
        if(mem[i]>-1) return mem[i];
        if(s[i]=='0') return mem[i] = 0;
        int res = num(i+1,s,mem);
        if(i<s.size()-1 && (s[i]=='1'||s[i]=='2'&&s[i+1]<'7')) res+=num(i+2,s,mem);
        return mem[i] = res;
    }  

Java: dp, O(n) time and space

int numDecodings(string s) {
        int n = s.size();
        vector<int> dp(n+1);
        dp[n] = 1;
        for(int i=n-1;i>=0;i--) {
            if(s[i]=='0') dp[i]=0;
            else {
                dp[i] = dp[i+1];
                if(i<n-1 && (s[i]=='1'||s[i]=='2'&&s[i+1]<'7')) dp[i]+=dp[i+2];
            }
        }
        return s.empty()? 0 : dp[0];   
    }

Java: dp, constance space 

int numDecodings(string s) {
        int p = 1, pp, n = s.size();
        for(int i=n-1;i>=0;i--) {
            int cur = s[i]=='0' ? 0 : p;
            if(i<n-1 && (s[i]=='1'||s[i]=='2'&&s[i+1]<'7')) cur+=pp;
            pp = p;
            p = cur;
        }
        return s.empty()? 0 : p;   
    }  

Java:

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:

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: wo

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        if len(s) == 0 or s[0] == '0':
            return 0
        
        dp = [0] * (len(s) + 1)
        dp[0] = 1
       
        for i in xrange(1, len(s) + 1):
            if s[i-1] != '0':
                dp[i] = dp[i-1]
            if i > 1 and (s[i-2] == '1' or (s[i-2] == '2' and int(s[i-1]) <= 6)):
                dp[i] += dp[i-2] 
                
        return dp[-1]     

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

C++:

class Solution {
public:
    int numDecodings(string s) {
        if (s.empty() || (s.size() > 1 && s[0] == '0')) return 0;
        vector<int> dp(s.size() + 1, 0);
        dp[0] = 1;
        for (int i = 1; i < dp.size(); ++i) {
            dp[i] = (s[i - 1] == '0') ? 0 : dp[i - 1];
            if (i > 1 && (s[i - 2] == '1' || (s[i - 2] == '2' && s[i - 1] <= '6'))) {
                dp[i] += dp[i - 2];
            }
        }
        return dp.back();
    }
};

C++:

class Solution {
public:
    int numDecodings(string s) {
        if (s.empty()) return 0;
        vector<int> dp(s.size() + 1, 0);
        dp[0] = 1;
        for (int i = 1; i < dp.size(); ++i) {
            if (s[i - 1] != '0') dp[i] += dp[i - 1];
            if (i >= 2 && s.substr(i - 2, 2) <= "26" && s.substr(i - 2, 2) >= "10") {
                dp[i] += dp[i - 2];
            }
        }
        return dp.back();
    }
};

C++:

class Solution {
public:
    int numDecodings(string s) {
        if (s.empty() || s.front() == '0') return 0;
        int c1 = 1, c2 = 1;
        for (int i = 1; i < s.size(); ++i) {
            if (s[i] == '0') c1 = 0;
            if (s[i - 1] == '1' || (s[i - 1] == '2' && s[i] <= '6')) {
                c1 = c1 + c2;
                c2 = c1 - c2;
            } else {
                c2 = c1;
            }
        }
        return c1;
    }
};

 

类似题目:

[LeetCode] 639. Decode Ways II 解码方法之二

 

All LeetCode Questions List 题目汇总

 

  

  

  

  

 

 

 

 

 

posted @ 2018-02-28 15:11  轻风舞动  阅读(494)  评论(0编辑  收藏  举报