[LintCode] Decode Ways 解码方法

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[LintCode] Decode Ways 解码方法

李博 bluemind 2017-12-14 07:44:00 浏览812
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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.

Example
Given encoded message 12, it could be decoded as AB (1 2) or L (12).
The number of ways decoding 12 is 2.

LeetCode上的原题,请参见我之前的博客Decode Ways

解法一:

class Solution {
public:
    /**
     * @param s a string,  encoded message
     * @return an integer, the number of ways decoding
     */
    int numDecodings(string& s) {
        if (s.empty()) return 0;
        int n = s.size();
        vector<int> dp(n + 2, 1);
        for (int i = 2; i < n + 2; ++i) {
            if (s[i - 2] == '0') dp[i] = 0;
            else dp[i] = dp[i - 1];
            if (i >= 3 && (s[i - 3] == '1' || (s[i - 3] == '2' && s[i - 2] <= '6'))) {
                dp[i] += dp[i -2];
            }
        }
        return dp.back();
    }
};

解法二:

class Solution {
public:
    /**
     * @param s a string,  encoded message
     * @return an integer, the number of ways decoding
     */
    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] >= '1' && s[i - 1] <= '9') 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();
    }
};

解法三:

class Solution {
public:
    /**
     * @param s a string,  encoded message
     * @return an integer, the number of ways decoding
     */
    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;
    }
};

 本文转自博客园Grandyang的博客,原文链接:解码方法[LintCode] Decode Ways ,如需转载请自行联系原博主。

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