String Matching（上交复试上机）

前言：

21考研，不论能否进复试记录一下准备路上写下的垃圾代码。本来啃《算法笔记》，但是感觉太多了做不完，改做王道机试指南。

题目描述：

Finding all occurrences of a pattern in a text is a problem that arises frequently in text-editing programs. Typically,the text is a document being edited,and the pattern searched for is a particular word supplied by the user. We assume that the text is an array T[1…n] of length n and that the pattern is an array P[1…m] of length m<=n.We further assume that the elements of P and T are all alphabets(∑={a,b…,z}).The character arrays P and T are often called strings of characters. We say that pattern P occurs with shift s in the text T if 0<=s<=n and T[s+1…s+m] = P[1…m](that is if T[s+j]=P[j],for 1<=j<=m). If P occurs with shift s in T,then we call s a valid shift;otherwise,we calls a invalid shift. Your task is to calculate the number of vald shifts for the given text T and p attern P.

输入描述

For each case, there are two strings T and P on a line,separated by a single space.You may assume both the length of T and P will not exceed 10^6.

输出描述:

You should output a number on a separate line,which indicates the number of valid shifts for the given text T and pattern P.

解答

#include<iostream>
#include<string>
#include<vector>
using namespace std;

const int MAXN = 1000 + 10;

void getnext(int next[], string t) {
    
    
	int k = -1, j = 0;
	next[0] = -1;
	while (j < t.length() -1 )
		if (k == -1 || t[k] == t[j])
			next[++j] = ++k;
		else
			k = next[k];
	return;
}

int KMP(string text,string t,int next[]) {
    
    
	int i = 0, j = 0;
	int count = 0;
	while (i < (int)(text.length()) && j < (int)(t.length())) {
    
    	
	//注意一定要加强制类型转换，length()返回的是无符号数，若不加当j=-1，比较的时候j被转为无符号数，会变得很大，导致bug
		if (j == -1 || text[i] == t[j]) {
    
    
			i++;
			j++;
		}
		else
			j = next[j];
		if (j == t.length()) {
    
    
			i = i - j + 1;
			j = 0;
			count++;
		}
	}
	return count;
}

int main() {
    
    
	string text, pattern;
	while (cin >> text >> pattern) {
    
    
		int next[MAXN];
		getnext(next, pattern);
		cout << KMP(text, pattern, next) << endl;
	}
	return 0;
}

可以用STL的find()实现非常方便，但是今天刚学完KMP算法，还是想手动实现一下。string的length()返回的是无符号数这点真的太坑了，想了很久想不明白，还是请教了大佬才了解。