【leetcode】Implement strStr

题目：

仅对这个题而言，用普通的方法，就是所谓的BF，相信大家都能够实现，区别只是谁的代码更简练，但是实现的算法都是一样的，这里写写KMP，至于kmp前前后后的来龙去脉，这里不做讲解，自认为还没这个能力，在走访过的博客论坛技术贴中，也未曾清楚的明白作者所言为何物，这里没有贬低的意思，每个作者将自己明白和理解的东西能无私的分享出来，都是非常棒的！

这里给 大家提供几个讲解KMP的链接，是能够跟着走一遍就能明白的那种。

1. 阮一峰：字符串的KMP算法

2.相似的英文博客：jBoxer

看了这两篇文章，如果不用你自己生成next表，估计现在你可以很快的写出kmp算法，并能讲解清楚。下面看看维基百科里给出的匹配算法和生成next表的伪代码。维基：Knuth-Morris-Pratt algorithm

//kmp search pseudo code

伪代码：

algorithm kmp_search:    input:        an array of characters, S (the text to be searched)        an array of characters, W (the word sought)    output:        an integer (the zero-based position in S at which W is found)    define variables:        an integer, m ← 0 (the beginning of the current match in S)        an integer, i ← 0 (the position of the current character in W)        an array of integers, T (the table, computed elsewhere)    while m + i < length(S) do        if W[i] = S[m + i] then            if i = length(W) - 1 then                return m            let i ← i + 1        else            if T[i] > -1 then                let i ← T[i], m ← m + i - T[i]            else                let i ← 0, m ← m + 1                (if we reach here, we have searched all of S unsuccessfully)    return the length of S

//build next table

algorithm kmp_table:    input:        an array of characters, W (the word to be analyzed)        an array of integers, T (the table to be filled)    output:        nothing (but during operation, it populates the table)    define variables:        an integer, pos ← 2 (the current position we are computing in T)        an integer, cnd ← 0 (the zero-based index in W of the next character of the current candidate substring)    (the first few values are fixed but different from what the algorithm might suggest)    let T[0] ← -1, T[1] ← 0    while pos < length(W) do        (first case: the substring continues)        if W[pos - 1] = W[cnd] then            let cnd ← cnd + 1, T[pos] ← cnd, pos ← pos + 1        (second case: it doesn't, but we can fall back)        else if cnd > 0 then            let cnd ← T[cnd]        (third case: we have run out of candidates.  Note cnd = 0)        else            let T[pos] ← 0, pos ← pos + 1

实现代码

按着伪代码我们试着实现一下。仅供参考。

void fill_table(string p, vector<int>& table){table[0] = -1;table[1] = 0;int index = 2;int cnt = 0;while (index < p.size()){if(p[cnt] == p[index - 1])p[index++] = ++cnt;else if(cnt > 0)cnt = table[cnt];elsetable[index++] = 0;}}int Compare(string s, string p){if(s.size() < p.size() || s.size() == 0 || p.size() == 0)return s.size();int m = 0;int i = 0;vector<int> table(p.size() + 1,0);fill_table(p, table);while (m + i < s.size()){if(i >= p.size()) return m;if(s[m + i] == p[i]) ++i;else{m += i - table[i];i = 0;}}return s.size();}