new algo

Exercise/Homework1/D KMP模式匹配算法.cpp

@@ -0,0 +1,71 @@
+#include <iostream>
+#include <string>
+#include <vector>
+using namespace std;
+vector<int> build_next(const string& p) {
+    int n = p.size();                 // 模式串长度
+    vector<int> nxt(n, 0);            // 存放 next 数组，初始化全 0
+
+    for (int i = 1, j = 0; i < n; i++) {
+        // --- step1: 如果当前字符不匹配，回退 j ---
+        while (j > 0 && p[i] != p[j]) 
+            j = nxt[j - 1];           // j 回退到上一个最长前后缀的长度，采用动态规划思想
+
+        // --- step2: 如果匹配成功，j 前进 ---
+        if (p[i] == p[j]) 
+            j++;
+
+        // --- step3: 记录当前前缀的 next 值 ---
+        nxt[i] = j;                   // 表示 p[0..i] 的最长公共前后缀长度
+    }
+
+    return nxt;
+}
+
+
+vector<int> next_arr(const string& x) {
+    int n = x.size();
+    vector<int> tmp(n, 0);
+
+    for (int i = 0; i < n; i++) {       
+        int j = 0;
+        for (int len = i; len > 0; len--) {
+            if (x.substr(0, len) == x.substr(i - len + 1, len)) {
+                j = len;
+                break;
+            }
+        }
+        tmp[i] = j;
+    }
+    return tmp;
+}
+
+
+int main(){
+    string mod, input;
+    cin >> input >> mod;
+    vector<int> next = build_next(mod);
+    int n = mod.size();
+    int pi = 0, pm = 0;
+    cout << next[n / 4] << " " << next[n / 2] << " " << next[3 * n / 4] << endl;
+    while (pi < input.size()) {
+        if (input[pi] == mod[pm]) {
+            pi++;
+            pm++;
+            if (pm == mod.size()) {
+                cout << pi - pm << " ";   // 匹配位置（起始下标）
+                return 0;
+                pm = next[pm - 1];       // 回退
+            }
+        } else {
+            if (pm > 0)
+                pm = next[pm - 1];       // 回退
+            else
+                pi++;
+        }
+    }
+    cout << -1 << endl;
+    return 0;
+}
+
+qwerababcabcabcabcdaabcabhlk