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  • leetcode 28

    题目描述:

    Implement strStr().

    Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

    分析:

    这个题目考察的是KMP算法,这个算法的核心是求出子串的自匹配数组,在失配情况下不回溯父串指针(其实是索引),只需将子串向左滑动即可,next数组就是记录了滑动的距离的数组,求next数组是这个算法的难点。

    我的第一个版本的代码如下:

    #include <iostream>
    #include <string>
    #include <vector>
    
    using namespace std;
    
    int strStr(string haystack, string needle);
    void getNext(vector<int> &ivec, const string & s);
    
    int main()
    {
        cout << "Hello world!" << endl;
        string hay("mississippi");
        string needle("issi");
        int res = strStr(hay, needle);
        cout << "the result of matching:" << res << endl;
        system("pause");
        return 0;
    }
    
    void getNext(vector<int> &ivec, const string & s)
    {
        int i = 0, j = -1;
        ivec[i] = j;
        while (i < s.size() - 1)
        {
            if (j == -1 || s[i] == s[j])
                ivec[++i] = ++j;
            else
                j = ivec[j];
        }
    }
    
    int strStr(string haystack, string needle)
    {
        if (needle.size() == 0)
            return 0;
        if (haystack.size() == 0)
            return -1;
    
        vector<int> ivec(needle.size(), 0);
        getNext(ivec, needle);
    
    
        for (auto i = ivec.begin(); i != ivec.end(); i++)
            cout << *i << endl;
    
    
        int i = 0, j = 0;
        cout << haystack.size() << "  " << needle.size() << endl;
    
        while( i < haystack.size() && j < needle.size() )
        //while (i != haystack.size() && j != needle.size())
        {
            cout << "before modified " << i << " " << j << endl;
            if (j == -1 || haystack[i] == needle[j])
            {
                cout << "what" << endl;
                ++i;
                ++j;
            }
            else
            {
                j = ivec[j];
            }
            cout << "after modified " << i << " " << j << endl;
    
        }
        if (j == needle.size())
            return i - j;
        else
            return -1;
    
    }

    这份代码我以为可以完美AC了,结果报了wrong answer,我很不理解,经过修改,就是将上一个版本中的while循环头换成下方紧邻的那一行就AC了,经过和大神的讨论,大神很快指出了我的错误所在,那就是C++的隐式类型转换!haystack.size()返回的是一个unsigned int类型,而i, j是int型,在运算时int会隐式转换为unsigned int,而-1转换为unsigned int型之后为4294967295(2的32次方减1)是一个非常大的数字!以前写循环不注意,现在吃了大亏!希望吃一堑长一智吧,以后写代码一定要非常注意类型的选择,注意int和unsigned int不能比较,如果必须比较一定将unsigned int 强制转换为long long类型再比较!最后完成版的代码如下:

    #include <iostream>
    #include <string>
    #include <vector>
    
    using namespace std;
    
    int strStr(string haystack, string needle);
    void getNext(vector<int> &ivec, const string & s);
    
    int main()
    {
        string hay("mississippi");
        string needle("issi");
        int res = strStr(hay, needle);
        cout << "the result of matching:" << res << endl;
        system("pause");
        return 0;
    }
    
    void getNext(vector<int> &ivec, const string & s)
    {
        int i = 0, j = -1;
        ivec[i] = j;
        while (i < s.size() - 1)
        {
            if (j == -1 || s[i] == s[j])
                ivec[++i] = ++j;
            else
                j = ivec[j];
        }
    }
    
    int strStr(string haystack, string needle)
    {
        if (needle.size() == 0)
            return 0;
        if (haystack.size() == 0)
            return -1;
    
        vector<int> ivec(needle.size(), 0);
        getNext(ivec, needle);
    
    
        for (auto i = ivec.begin(); i != ivec.end(); i++)
            cout << *i << endl;
    
    
        int i = 0, j = 0;
        long long haystack_size = haystack.size();
        long long needle_size = needle.size();
    
        while( i < haystack_size && j < needle_size )
        {
            cout << "i and j before modified " << i << " " << j << endl;
            if (j == -1 || haystack[i] == needle[j])
            {
                cout << "what" << endl;
                ++i;
                ++j;
            }
            else
            {
                j = ivec[j];
            }
            cout << "i and j after modified " << i << " " << j << endl;
            haystack_size = haystack.size();
            needle_size = needle.size();
    
        }
        if (j == needle.size())
            return i - j;
        else
            return -1;
    
    }
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  • 原文地址:https://www.cnblogs.com/maizi-1993/p/5929100.html
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