新增 LeetCode 28 找出字符串中第一个匹配项的下标的 C++ 实现，使用 KMP 算法优化查找效率

Author: tkzzzzzz6

acwing/高校试题/北航/3526. 素数.cpp

@@ -0,0 +1,42 @@
+/*
+ * @Author: tkzzzzzz6
+ * @Date: 2026-04-13 11:23:16
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-04-14 01:42:29
+ */
+/*
+* @acwing app=acwing.cn id=3529 lang=C++
+*
+* 3526. 素数
+*/
+
+// @acwing code start
+#include<iostream>
+
+using namespace std;
+
+bool is_prime(int x){
+    if(x < 2 || x % 2 == 0)return false;
+    for(int i = 2;i<=x/i;++i){
+        if(x % i == 0)return false;
+    }
+    return true;
+}
+
+int main(){
+    int n;
+    while(cin >> n){
+        int cnt = 0;
+        for(int i = 2;i<n;i = (i / 10 + 1)*10 + 1){
+            if(is_prime(i)){
+                ++cnt;
+                cout << i << ' ';
+            }
+        }
+        if(!cnt) cout << -1;
+        cout << endl;
+    }
+    return 0;
+}
+
+// @acwing code end

leetcode/Study Plan/28. 找出字符串中第一个匹配项的下标/1.cpp

@@ -1,12 +1,67 @@
 /*
  * @Author: tkzzzzzz6
- * @Date: 2026-04-14 01:07:56
+ * @Date: 2026-04-14 01:08:25
  * @LastEditors: tkzzzzzz6
- * @LastEditTime: 2026-04-14 01:08:01
+ * @LastEditTime: 2026-04-14 01:21:50
  */
+/*
+ * @lc app=leetcode.cn id=28 lang=cpp
+ * @lcpr version=30204
+ *
+ * [28] 找出字符串中第一个匹配项的下标
+ */
+
+
+// @lcpr-template-start
+using namespace std;
+#include <algorithm>
+#include <array>
+#include <bitset>
+#include <climits>
+#include <deque>
+#include <functional>
+#include <iostream>
+#include <list>
+#include <queue>
+#include <stack>
+#include <tuple>
+#include <unordered_map>
+#include <unordered_set>
+#include <utility>
+#include <vector>
+// @lcpr-template-end
+// @lc code=start
 class Solution {
 public:
     int strStr(string haystack, string needle) {
-        int n = haystack.size(),m = needle
+        int m = haystack.size(),n = needle.size();
+        if(n == 0)return -1;
+        string s = needle + '#' + haystack;
+        vector<int> pi(s.size());
+        for(int i = 1;i<s.size();++i){
+            int len = pi[i - 1];
+            while(len && s[i] != s[len]){
+                len = pi[len-1];
+            }
+            if(s[i] == s[len]){
+                pi[i] = len + 1;
+                if(pi[i] == n)return i-2*n;
+            }
+        }
+        return -1;
     }
 };
+// @lc code=end
+
+
+
+/*
+// @lcpr case=start
+// "sadbutsad"\n"sad"\n
+// @lcpr case=end
+
+// @lcpr case=start
+// "leetcode"\n"leeto"\n
+// @lcpr case=end
+
+ */