笔者介绍：姜雪伟，IT公司技术合伙人，IT高级讲师，CSDN社区专家，特邀编辑，畅销书作者，已出版书籍：《手把手教你架构3D游戏引擎》电子工业出版社和《Unity3D实战核心技术详解》电子工业出版社等。
CSDN视频网址：http://edu.csdn.net/lecturer/144
算法在各个领域都有非常广泛的应用，尤其现在比较流行的人工智能，深度学习以及搜索算法等等，当然游戏开发领域也需要算法的支撑，只是这些算法被一些库已封装好，不需要开发者重新编写，但是这不等于大家就可以不用学习算法了。本篇博客给读者介绍一种KMP模型搜索算法，学过数据结构的读者知道，搜索算法很多的，比如二分查找，二叉树搜索等等。
KMP算法是一种改进的字符串匹配算法，由D.E.Knuth，J.H.Morris和V.R.Pratt同时发现，因此人们称它为克努特——莫里斯——普拉特操作（简称KMP算法）。KMP算法的关键是利用匹配失败后的信息，尽量减少模式串与主串的匹配次数以达到快速匹配的目的。具体实现就是实现一个next()函数，函数本身包含了模式串的局部匹配信息。时间复杂度O(m+n)。网上很多关于KMP算法的解释，在这里就不多说了，直接通过案例给读者分享，通过代码的分析可以了解其算法的运行原理。
案例如下所示：
输入:
txt[] = "THIS IS A TEST TEXT" pat[] = "TEST"
输出:
Pattern found at index 10输入:
txt[] = "AABAACAADAABAAABAA" pat[] = "AABA"输出:
Pattern found at index 0 Pattern found at index 9 Pattern found at index 13#include<stdio.h>#include<string.h>#include<stdlib.h> void computeLPSArray(char *pat, int M, int *lps); void KMPSearch(char *pat, char *txt){ int M = strlen(pat); int N = strlen(txt); // create lps[] that will hold the longest prefix suffix // values for pattern int *lps = (int *)malloc(sizeof(int)*M); int j = 0; // index for pat[] // Preprocess the pattern (calculate lps[] array) computeLPSArray(pat, M, lps); int i = 0; // index for txt[] while (i < N) { if (pat[j] == txt[i]) { j++; i++; } if (j == M) { printf("Found pattern at index %d \n", i-j); j = lps[j-1]; } // mismatch after j matches else if (i < N && pat[j] != txt[i]) { // Do not match lps[0..lps[j-1]] characters, // they will match anyway if (j != 0) j = lps[j-1]; else i = i+1; } } free(lps); // to avoid memory leak} void computeLPSArray(char *pat, int M, int *lps){ int len = 0; // length of the previous longest prefix suffix int i; lps[0] = 0; // lps[0] is always 0 i = 1; // the loop calculates lps[i] for i = 1 to M-1 while (i < M) { if (pat[i] == pat[len]) { len++; lps[i] = len; i++; } else // (pat[i] != pat[len]) { if (len != 0) { // This is tricky. Consider the example // AAACAAAA and i = 7. len = lps[len-1]; // Also, note that we do not increment i here } else // if (len == 0) { lps[i] = 0; i++; } } }} // Driver program to test above functionint main(){ char *txt = "ABABDABACDABABCABAB"; char *pat = "ABABCABAB"; KMPSearch(pat, txt); return 0;}
现在Python语言也是比较流行的，再用Python实现一遍代码如下所示：
# Python program for KMP Algorithmdef KMPSearch(pat, txt): M = len(pat) N = len(txt) # create lps[] that will hold the longest prefix suffix # values for pattern lps = [0]*M j = 0 # index for pat[] # Preprocess the pattern (calculate lps[] array) computeLPSArray(pat, M, lps) i = 0 # index for txt[] while i < N: if pat[j] == txt[i]: i += 1 j += 1 if j == M: print "Found pattern at index " + str(i-j) j = lps[j-1] # mismatch after j matches elif i < N and pat[j] != txt[i]: # Do not match lps[0..lps[j-1]] characters, # they will match anyway if j != 0: j = lps[j-1] else: i += 1 def computeLPSArray(pat, M, lps): len = 0 # length of the previous longest prefix suffix lps[0] # lps[0] is always 0 i = 1 # the loop calculates lps[i] for i = 1 to M-1 while i < M: if pat[i]==pat[len]: len += 1 lps[i] = len i += 1 else: # This is tricky. Consider the example. # AAACAAAA and i = 7. The idea is similar # to search step. if len != 0: len = lps[len-1] # Also, note that we do not increment i here else: lps[i] = 0 i += 1 txt = "ABABDABACDABABCABAB"pat = "ABABCABAB"KMPSearch(pat, txt)
当然Java语言也是必不可少的，Java编写的代码如下所示：
class KMP_String_Matching{ void KMPSearch(String pat, String txt) { int M = pat.length(); int N = txt.length(); // create lps[] that will hold the longest // prefix suffix values for pattern int lps[] = new int[M]; int j = 0; // index for pat[] // Preprocess the pattern (calculate lps[] // array) computeLPSArray(pat,M,lps); int i = 0; // index for txt[] while (i < N) { if (pat.charAt(j) == txt.charAt(i)) { j++; i++; } if (j == M) { System.out.println("Found pattern "+ "at index " + (i-j)); j = lps[j-1]; } // mismatch after j matches else if (i < N && pat.charAt(j) != txt.charAt(i)) { // Do not match lps[0..lps[j-1]] characters, // they will match anyway if (j != 0) j = lps[j-1]; else i = i+1; } } } void computeLPSArray(String pat, int M, int lps[]) { // length of the previous longest prefix suffix int len = 0; int i = 1; lps[0] = 0; // lps[0] is always 0 // the loop calculates lps[i] for i = 1 to M-1 while (i < M) { if (pat.charAt(i) == pat.charAt(len)) { len++; lps[i] = len; i++; } else // (pat[i] != pat[len]) { // This is tricky. Consider the example. // AAACAAAA and i = 7. The idea is similar // to search step. if (len != 0) { len = lps[len-1]; // Also, note that we do not increment // i here } else // if (len == 0) { lps[i] = len; i++; } } } } // Driver program to test above function public static void main(String args[]) { String txt = "ABABDABACDABABCABAB"; String pat = "ABABCABAB"; new KMP_String_Matching().KMPSearch(pat,txt); }}
算法也是检验开发者处理问题的能力，在业余时间没事的时候，可以编写一下就当放松了。。。。。。。