BruteForceStringMatching

//Recall the string-matching problem introduced before: given a string of//n characters called text and a string of m characters(m <= n) called the//pattern, find a substring of the text that matches the pattern. To put//it more precisely, we want to find the i - the index of the leftmost character//of the first matching substring in the text.//If matches other than the first one need to be found, a string-matching algorithm//can simply continue working until the entire text is exhausted.//A brute-force algorithm for the string-matching problem is quite obvious://align the pattern against the first m characters of the text and start matching//the corresponding pairs of characters from left to right until either all the//m pairs of the characters match(then the algorithm can stop) or a mismatching//pair is encountered. In the latter case, shift the patter on position to the//right and resume the character comparisons, starting again with the first//character of the pattern and its counterpart in the text. Note that the last//position in the text that can still be a beginning of a matching substring//is n - m(provided the text positions are indexed from 0 to n-1). Beyond that//position, there are not enough characters to match the entire pattern; hence,//the algorithm need not make any comparisons there.package mainimport (    "fmt")func BruteForceStringMatching(text, pattern []rune) int {    n := len(text)    m := len(pattern)    for i := 0; i <= n-m; i++ {        for j := 0; j < m && pattern[j] == text[i+j]; {            j++            if j == m {                return i            }        }    }    return -1}func main() {    text := "暴力法解决字符串匹配问题，字符串匹配问题用暴力法"    pattern := "暴力"    fmt.Println(BruteForceStringMatching([]rune(text), []rune(pattern)))    text2 := []rune("abedefgabdedsdfsesdffses")    patter2 := []rune("ses")    fmt.Println(BruteForceStringMatching(text2, patter2))}