给定文本字符串和模式字符串，在字符串中查找模式的所有匹配项。
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几种模式搜索算法( 国民党 , 拉宾·卡普 , 朴素算法 , 有限自动机 )已经讨论过，可用于此检查。 这里我们将讨论基于后缀树的算法。
在1 圣 后缀树应用程序( 子串检查 )，我们看到了如何检查给定的模式是否是文本的子字符串。建议通过 子串检查 1. 圣 . 在本文中，我们将进一步讨论同样的问题。如果模式是文本的子字符串，那么我们将找到文本中模式上的所有位置。
作为先决条件，我们必须知道如何以一种或另一种方式构建后缀树。 在这里，我们将使用Ukkonen的算法构建后缀树，如下所述： Ukkonen后缀树构造——第1部分 Ukkonen后缀树构造——第2部分 Ukkonen后缀树构造——第3部分 Ukkonen后缀树构造——第4部分 Ukkonen后缀树构造——第5部分 Ukkonen后缀树构造——第6部分
让我们看看下图： 
这是字符串“abcabxabcd$”的后缀树，显示后缀索引和边标签索引（开始、结束）。边上显示的（子）字符串值仅用于解释目的。我们从不在树中存储路径标签字符串。 路径的后缀索引表示该路径上的子字符串（从根开始）的索引。 在上面的树中考虑后缀索引7的“BCD $”路径。它表明子字符串b、bc、bcd、bcd$在字符串的索引7处。 类似地，带有后缀索引4的路径“bxabcd$”表明子字符串b、bx、bxa、bxab、bxabc、bxabcd、bxabcd$位于索引4处。 类似地，带有后缀索引1的路径“bcabxabcd$”表明子字符串b、bc、bca、bcab、bcabx、bcabxa、bcabxab、bcabxabc、bcabxabcd、bcabxabcd$位于索引1处。
如果我们把以上三条路放在一起看，我们可以看到：
- 子串“b”位于索引1、4和7处
- 子串“bc”位于索引1和7处
通过以上解释，我们应该能够看到以下内容：
- 子串“ab”位于索引0、3和6处
- 子字符串“abc”位于索引0和6处
- 子串“c”位于索引2和8处
- 子字符串“xab”位于索引5处
- 子字符串“d”位于索引9处
- 子字符串“cd”位于索引8处
….. …..
你能看到如何在一个字符串中找到所有出现的图案吗？
- 1. 圣 首先，检查给定的模式是否真的存在于字符串中（就像我们在 子串检查 ).为此，根据模式遍历后缀树。
- 如果在后缀树中找到模式（不要从树上掉下来），则遍历该点下方的子树，找到叶节点上的所有后缀索引。所有这些后缀索引都将是字符串中的模式索引
// A C program to implement Ukkonen's Suffix Tree Construction // And find all locations of a pattern in string #include <stdio.h> #include <string.h> #include <stdlib.h> #define MAX_CHAR 256 struct SuffixTreeNode { struct SuffixTreeNode *children[MAX_CHAR]; //pointer to other node via suffix link struct SuffixTreeNode *suffixLink; /*(start, end) interval specifies the edge, by which the node is connected to its parent node. Each edge will connect two nodes, one parent and one child, and (start, end) interval of a given edge will be stored in the child node. Let's say there are two nods A and B connected by an edge with indices (5, 8) then this indices (5, 8) will be stored in node B. */ int start; int *end; /*for leaf nodes, it stores the index of suffix for the path from root to leaf*/ int suffixIndex; }; typedef struct SuffixTreeNode Node; char text[100]; //Input string Node *root = NULL; //Pointer to root node /*lastNewNode will point to the newly created internal node, waiting for it's suffix link to be set, which might get a new suffix link (other than root) in next extension of same phase. lastNewNode will be set to NULL when last newly created internal node (if there is any) got it's suffix link reset to new internal node created in next extension of same phase. */ Node *lastNewNode = NULL; Node *activeNode = NULL; /*activeEdge is represented as an input string character index (not the character itself)*/ int activeEdge = -1; int activeLength = 0; // remainingSuffixCount tells how many suffixes yet to // be added in tree int remainingSuffixCount = 0; int leafEnd = -1; int *rootEnd = NULL; int *splitEnd = NULL; int size = -1; //Length of input string Node *newNode( int start, int *end) { Node *node =(Node*) malloc ( sizeof (Node)); int i; for (i = 0; i < MAX_CHAR; i++) node->children[i] = NULL; /*For root node, suffixLink will be set to NULL For internal nodes, suffixLink will be set to root by default in current extension and may change in next extension*/ node->suffixLink = root; node->start = start; node->end = end; /*suffixIndex will be set to -1 by default and actual suffix index will be set later for leaves at the end of all phases*/ node->suffixIndex = -1; return node; } int edgeLength(Node *n) { if (n == root) return 0; return *(n->end) - (n->start) + 1; } int walkDown(Node *currNode) { /*activePoint change for walk down (APCFWD) using Skip/Count Trick (Trick 1). If activeLength is greater than current edge length, set next internal node as activeNode and adjust activeEdge and activeLength accordingly to represent same activePoint*/ if (activeLength >= edgeLength(currNode)) { activeEdge += edgeLength(currNode); activeLength -= edgeLength(currNode); activeNode = currNode; return 1; } return 0; } void extendSuffixTree( int pos) { /*Extension Rule 1, this takes care of extending all leaves created so far in tree*/ leafEnd = pos; /*Increment remainingSuffixCount indicating that a new suffix added to the list of suffixes yet to be added in tree*/ remainingSuffixCount++; /*set lastNewNode to NULL while starting a new phase, indicating there is no internal node waiting for it's suffix link reset in current phase*/ lastNewNode = NULL; //Add all suffixes (yet to be added) one by one in tree while (remainingSuffixCount > 0) { if (activeLength == 0) activeEdge = pos; //APCFALZ // There is no outgoing edge starting with // activeEdge from activeNode if (activeNode->children] == NULL) { //Extension Rule 2 (A new leaf edge gets created) activeNode->children] = newNode(pos, &leafEnd); /*A new leaf edge is created in above line starting from an existing node (the current activeNode), and if there is any internal node waiting for its suffix link get reset, point the suffix link from that last internal node to current activeNode. Then set lastNewNode to NULL indicating no more node waiting for suffix link reset.*/ if (lastNewNode != NULL) { lastNewNode->suffixLink = activeNode; lastNewNode = NULL; } } // There is an outgoing edge starting with activeEdge // from activeNode else { // Get the next node at the end of edge starting // with activeEdge Node *next = activeNode->children]; if (walkDown(next)) //Do walkdown { //Start from next node (the new activeNode) continue ; } /*Extension Rule 3 (current character being processed is already on the edge)*/ if (text[next->start + activeLength] == text[pos]) { //If a newly created node waiting for it's //suffix link to be set, then set suffix link //of that waiting node to current active node if (lastNewNode != NULL && activeNode != root) { lastNewNode->suffixLink = activeNode; lastNewNode = NULL; } //APCFER3 activeLength++; /*STOP all further processing in this phase and move on to next phase*/ break ; } /*We will be here when activePoint is in middle of the edge being traversed and current character being processed is not on the edge (we fall off the tree). In this case, we add a new internal node and a new leaf edge going out of that new node. This is Extension Rule 2, where a new leaf edge and a new internal node get created*/ splitEnd = ( int *) malloc ( sizeof ( int )); *splitEnd = next->start + activeLength - 1; //New internal node Node *split = newNode(next->start, splitEnd); activeNode->children] = split; //New leaf coming out of new internal node split->children] = newNode(pos, &leafEnd); next->start += activeLength; split->children] = next; /*We got a new internal node here. If there is any internal node created in last extensions of same phase which is still waiting for it's suffix link reset, do it now.*/ if (lastNewNode != NULL) { /*suffixLink of lastNewNode points to current newly created internal node*/ lastNewNode->suffixLink = split; } /*Make the current newly created internal node waiting for it's suffix link reset (which is pointing to root at present). If we come across any other internal node (existing or newly created) in next extension of same phase, when a new leaf edge gets added (i.e. when Extension Rule 2 applies is any of the next extension of same phase) at that point, suffixLink of this node will point to that internal node.*/ lastNewNode = split; } /* One suffix got added in tree, decrement the count of suffixes yet to be added.*/ remainingSuffixCount--; if (activeNode == root && activeLength > 0) //APCFER2C1 { activeLength--; activeEdge = pos - remainingSuffixCount + 1; } else if (activeNode != root) //APCFER2C2 { activeNode = activeNode->suffixLink; } } } void print( int i, int j) { int k; for (k=i; k<=j; k++) printf ( "%c" , text[k]); } //Print the suffix tree as well along with setting suffix index //So tree will be printed in DFS manner //Each edge along with it's suffix index will be printed void setSuffixIndexByDFS(Node *n, int labelHeight) { if (n == NULL) return ; if (n->start != -1) //A non-root node { //Print the label on edge from parent to current node //Uncomment below line to print suffix tree // print(n->start, *(n->end)); } int leaf = 1; int i; for (i = 0; i < MAX_CHAR; i++) { if (n->children[i] != NULL) { //Uncomment below two lines to print suffix index // if (leaf == 1 && n->start != -1) // printf(" [%d]", n->suffixIndex); //Current node is not a leaf as it has outgoing //edges from it. leaf = 0; setSuffixIndexByDFS(n->children[i], labelHeight + edgeLength(n->children[i])); } } if (leaf == 1) { n->suffixIndex = size - labelHeight; //Uncomment below line to print suffix index //printf(" [%d]", n->suffixIndex); } } void freeSuffixTreeByPostOrder(Node *n) { if (n == NULL) return ; int i; for (i = 0; i < MAX_CHAR; i++) { if (n->children[i] != NULL) { freeSuffixTreeByPostOrder(n->children[i]); } } if (n->suffixIndex == -1) free (n->end); free (n); } /*Build the suffix tree and print the edge labels along with suffixIndex. suffixIndex for leaf edges will be >= 0 and for non-leaf edges will be -1*/ void buildSuffixTree() { size = strlen (text); int i; rootEnd = ( int *) malloc ( sizeof ( int )); *rootEnd = - 1; /*Root is a special node with start and end indices as -1, as it has no parent from where an edge comes to root*/ root = newNode(-1, rootEnd); activeNode = root; //First activeNode will be root for (i=0; i<size; i++) extendSuffixTree(i); int labelHeight = 0; setSuffixIndexByDFS(root, labelHeight); } int traverseEdge( char *str, int idx, int start, int end) { int k = 0; //Traverse the edge with character by character matching for (k=start; k<=end && str[idx] != ' ' ; k++, idx++) { if (text[k] != str[idx]) return -1; // mo match } if (str[idx] == ' ' ) return 1; // match return 0; // more characters yet to match } int doTraversalToCountLeaf(Node *n) { if (n == NULL) return 0; if (n->suffixIndex > -1) { printf ( "Found at position: %d" , n->suffixIndex); return 1; } int count = 0; int i = 0; for (i = 0; i < MAX_CHAR; i++) { if (n->children[i] != NULL) { count += doTraversalToCountLeaf(n->children[i]); } } return count; } int countLeaf(Node *n) { if (n == NULL) return 0; return doTraversalToCountLeaf(n); } int doTraversal(Node *n, char * str, int idx) { if (n == NULL) { return -1; // no match } int res = -1; //If node n is not root node, then traverse edge //from node n's parent to node n. if (n->start != -1) { res = traverseEdge(str, idx, n->start, *(n->end)); if (res == -1) //no match return -1; if (res == 1) //match { if (n->suffixIndex > -1) printf ( "substring count: 1 and position: %d" , n->suffixIndex); else printf ( "substring count: %d" , countLeaf(n)); return 1; } } //Get the character index to search idx = idx + edgeLength(n); //If there is an edge from node n going out //with current character str[idx], traverse that edge if (n->children[str[idx]] != NULL) return doTraversal(n->children[str[idx]], str, idx); else return -1; // no match } void checkForSubString( char * str) { int res = doTraversal(root, str, 0); if (res == 1) printf ( "Pattern <%s> is a Substring" , str); else printf ( "Pattern <%s> is NOT a Substring" , str); } // driver program to test above functions int main( int argc, char *argv[]) { strcpy (text, "GEEKSFORGEEKS$" ); buildSuffixTree(); printf ( "Text: GEEKSFORGEEKS, Pattern to search: GEEKS" ); checkForSubString( "GEEKS" ); printf ( "Text: GEEKSFORGEEKS, Pattern to search: GEEK1" ); checkForSubString( "GEEK1" ); printf ( "Text: GEEKSFORGEEKS, Pattern to search: FOR" ); checkForSubString( "FOR" ); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); strcpy (text, "AABAACAADAABAAABAA$" ); buildSuffixTree(); printf ( "Text: AABAACAADAABAAABAA, Pattern to search: AABA" ); checkForSubString( "AABA" ); printf ( "Text: AABAACAADAABAAABAA, Pattern to search: AA" ); checkForSubString( "AA" ); printf ( "Text: AABAACAADAABAAABAA, Pattern to search: AAE" ); checkForSubString( "AAE" ); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); strcpy (text, "AAAAAAAAA$" ); buildSuffixTree(); printf ( "Text: AAAAAAAAA, Pattern to search: AAAA" ); checkForSubString( "AAAA" ); printf ( "Text: AAAAAAAAA, Pattern to search: AA" ); checkForSubString( "AA" ); printf ( "Text: AAAAAAAAA, Pattern to search: A" ); checkForSubString( "A" ); printf ( "Text: AAAAAAAAA, Pattern to search: AB" ); checkForSubString( "AB" ); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); return 0; } |
输出：
Text: GEEKSFORGEEKS, Pattern to search: GEEKS Found at position: 8 Found at position: 0 substring count: 2 Pattern <GEEKS> is a Substring Text: GEEKSFORGEEKS, Pattern to search: GEEK1 Pattern <GEEK1> is NOT a Substring Text: GEEKSFORGEEKS, Pattern to search: FOR substring count: 1 and position: 5 Pattern <FOR> is a Substring Text: AABAACAADAABAAABAA, Pattern to search: AABA Found at position: 13 Found at position: 9 Found at position: 0 substring count: 3 Pattern <AABA> is a Substring Text: AABAACAADAABAAABAA, Pattern to search: AA Found at position: 16 Found at position: 12 Found at position: 13 Found at position: 9 Found at position: 0 Found at position: 3 Found at position: 6 substring count: 7 Pattern <AA> is a Substring Text: AABAACAADAABAAABAA, Pattern to search: AAE Pattern <AAE> is NOT a Substring Text: AAAAAAAAA, Pattern to search: AAAA Found at position: 5 Found at position: 4 Found at position: 3 Found at position: 2 Found at position: 1 Found at position: 0 substring count: 6 Pattern <AAAA> is a Substring Text: AAAAAAAAA, Pattern to search: AA Found at position: 7 Found at position: 6 Found at position: 5 Found at position: 4 Found at position: 3 Found at position: 2 Found at position: 1 Found at position: 0 substring count: 8 Pattern <AA> is a Substring Text: AAAAAAAAA, Pattern to search: A Found at position: 8 Found at position: 7 Found at position: 6 Found at position: 5 Found at position: 4 Found at position: 3 Found at position: 2 Found at position: 1 Found at position: 0 substring count: 9 Pattern <A> is a Substring Text: AAAAAAAAA, Pattern to search: AB Pattern <AB> is NOT a Substring
Ukkonen的后缀树构造需要O（N）时间和空间来为长度为N的字符串构建后缀树，之后，子字符串检查的遍历需要对长度为M的模式进行O（M），然后如果该模式出现Z次，则需要O（Z）来找到所有这些Z次出现的索引。 总体模式复杂性是线性的：O（M+Z）。
更详细的分析 长度为N的字符串的后缀树中会有多少内部节点？？ 回答：N-1（为什么？？） 长度为N的字符串中将有N个后缀。 每个后缀将有一片叶子。 所以一个长度为N的后缀树将有N片叶子。 由于每个内部节点至少有2个子节点，一个N叶后缀树最多有N-1个内部节点。 如果一个模式在字符串中出现Z次，这意味着它将是Z后缀的一部分，因此在模式匹配在树中结束的点（内部节点和中间边）下方将有Z叶，因此在该点下方有Z叶的子树将有Z-1内部节点。一棵有Z片叶子的树可以在O（Z）时间内穿过。 总体模式复杂性是线性的：O（M+Z）。 对于给定的图案，Z（出现次数）可以是atmost N。 因此，最坏情况下的复杂度可能是：O（M+N），如果Z接近/等于N（一个有N个节点的树遍历需要O（N）时间）。
后续问题：
- 检查模式是否是文本的前缀？
- 检查模式是否为文本的后缀？
我们发表了以下更多关于后缀树应用程序的文章：
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