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  • 44. Decode Ways && Gray Code

    Decode Ways

    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.

    思想: 动态规划,分析:

    12121212:  前面为1 或者 (前面是2 且 当前值 < 7) : f(n) = f(n-1) +f (n-2);

    1202:  当前值为 0,前面是 1 或 2, f(n) = f(n-2). 否则,返回 0。

    其他情况, f(n) = f(n-1);

    class Solution {
    public:
        int numDecodings(string s) {
            int n = s.size();
            if(n == 0 || s[0] == '0') return 0;
            vector<int> f(n+1);
            f[0] = f[1] = 1;
            for(int index = 2; index <= n; ++index) {
                if('0' == s[index-1])  { 
                    if('1' == s[index-2] || '2' == s[index-2])  f[index] = f[index-2]; 
                    else return 0; 
                } 
                else if('1' == s[index-2] || ( '2' == s[index-2] && s[index-1] < '7'))  f[index] = f[index-1] + f[index-2];
                else f[index] = f[index-1];
            }
            return f[n];
        }
    };
    

    Gray Code

    The gray code is a binary numeral system where two successive values differ in only one bit.

    Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.

    For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:

    00 - 0
    01 - 1
    11 - 3
    10 - 2
    

    Note: For a given n, a gray code sequence is not uniquely defined.

    For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.

    For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.

    思想: 方法比较巧妙。 每次改变循环一轮,只改变最高位  0 ->1。

    class Solution {
    public:
        vector<int> grayCode(int n) {
            vector<int> vec(1, 0);
            for(int i = 0; i < n; ++i) {
                int v = 1 << i;
                for(int j = vec.size()-1; j >= 0; --j)
                    vec.push_back(vec[j]+v);
            }
            return vec;
        }
    };
    
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  • 原文地址:https://www.cnblogs.com/liyangguang1988/p/3940338.html
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