diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md
index 5fb28c29..bc73f6da 100644
--- a/problems/0150.逆波兰表达式求值.md
+++ b/problems/0150.逆波兰表达式求值.md
@@ -489,6 +489,67 @@ impl Solution {
 }
 ```
 
+### C:
+
+```c
+int str_to_int(char *str) {
+    // string转integer
+    int num = 0, tens = 1;
+    for (int i = strlen(str) - 1; i >= 0; i--) {
+        if (str[i] == '-') {
+            num *= -1;
+            break;
+        }
+        num += (str[i] - '0') * tens;
+        tens *= 10;
+    }
+    return num;
+}
+
+int evalRPN(char** tokens, int tokensSize) {
+
+    int *stack = (int *)malloc(tokensSize * sizeof(int));
+    assert(stack);
+    int stackTop = 0;
+
+    for (int i = 0; i < tokensSize; i++) {
+        char symbol = (tokens[i])[0];
+        if (symbol < '0' && (tokens[i])[1] == '\0') {
+
+            // pop两个数字
+            int num1 = stack[--stackTop];
+            int num2 = stack[--stackTop];
+
+            // 计算结果
+            int result;
+            if (symbol == '+') {
+                result = num1 + num2;
+            } else if (symbol == '-') {
+                result = num2 - num1;
+            } else if (symbol == '/') {
+                result = num2 / num1;
+            } else {
+                result = num1 * num2;
+            }
+
+            // push回stack
+            stack[stackTop++] = result;
+
+        } else {
+
+            // push数字进stack
+            int num = str_to_int(tokens[i]);
+            stack[stackTop++] = num;
+            
+        }
+    }
+
+    int result = stack[0];
+    free(stack);
+    return result;
+}
+```
+
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md
index f0fe3a3c..73d9db1b 100644
--- a/problems/0225.用队列实现栈.md
+++ b/problems/0225.用队列实现栈.md
@@ -1277,6 +1277,95 @@ impl MyStack {
 }
 ```
 
+### C:
+
+> C：单队列
+
+```c
+typedef struct Node {
+    int val;
+    struct Node *next;
+} Node_t;
+
+// 用单向链表实现queue
+typedef struct {
+    Node_t *head;
+    Node_t *foot;
+    int size;
+} MyStack;
+
+MyStack* myStackCreate() {
+    MyStack *obj = (MyStack *)malloc(sizeof(MyStack));
+    assert(obj);
+    obj->head = NULL;
+    obj->foot = NULL;
+    obj->size = 0;
+    return obj;
+}
+
+void myStackPush(MyStack* obj, int x) {
+
+    Node_t *temp = (Node_t *)malloc(sizeof(Node_t));
+    assert(temp);
+    temp->val = x;
+    temp->next = NULL;
+
+    // 添加至queue末尾
+    if (obj->foot) {
+        obj->foot->next = temp;
+    } else {
+        obj->head = temp;
+    }
+    obj->foot = temp;
+    obj->size++;
+}
+
+int myStackPop(MyStack* obj) {
+
+    // 获取末尾元素
+    int target = obj->foot->val;
+
+    if (obj->head == obj->foot) {
+        free(obj->foot);
+        obj->head = NULL;
+        obj->foot = NULL;
+    } else {
+        
+        Node_t *prev = obj->head;
+        // 移动至queue尾部节点前一个节点
+        while (prev->next != obj->foot) {
+            prev = prev->next;
+        }
+
+        free(obj->foot);
+        obj->foot = prev;
+        obj->foot->next = NULL;
+    }
+
+    obj->size--;
+    return target;
+}
+
+int myStackTop(MyStack* obj) {
+    return obj->foot->val;
+}
+
+bool myStackEmpty(MyStack* obj) {
+    return obj->size == 0;
+}
+
+void myStackFree(MyStack* obj) {
+    Node_t *curr = obj->head;
+    while (curr != NULL) {
+        Node_t *temp = curr->next;
+        free(curr);
+        curr = temp;
+    }
+    free(obj);
+}
+
+```
+
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md
index caa24d8d..9bb3494d 100644
--- a/problems/0239.滑动窗口最大值.md
+++ b/problems/0239.滑动窗口最大值.md
@@ -890,6 +890,38 @@ public:
 };
 ```
 
+### C
+
+```c
+int* maxSlidingWindow(int* nums, int numsSize, int k, int* returnSize) {
+    *returnSize = numsSize - k + 1;
+    int *res = (int*)malloc((*returnSize) * sizeof(int));
+    assert(res);
+    int *deque = (int*)malloc(numsSize * sizeof(int));
+    assert(deque);
+    int front = 0, rear = 0, idx = 0;
+
+    for (int i  = 0 ; i < numsSize ; i++) {
+        while (front < rear && deque[front] <= i - k) {
+            front++;
+        }
+
+        while (front < rear && nums[deque[rear - 1]] <= nums[i]) {
+            rear--;
+        }
+
+        deque[rear++] = i;
+
+        if (i >= k - 1) {
+            res[idx++] = nums[deque[front]];
+        }
+    }       
+
+    return res;
+}
+
+```
+
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index 2be8922b..de0e6e4d 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -879,6 +879,52 @@ public int[] GetNext(string s)
 }
 ```
 
+### C
+
+```c
+// 前缀表不减一
+int *build_next(char* s, int len) {
+
+    int *next = (int *)malloc(len * sizeof(int));
+    assert(next);
+
+    // 初始化前缀表
+    next[0] = 0;
+
+    // 构建前缀表表
+    int i = 1, j = 0;
+    while (i < len) {
+        if (s[i] == s[j]) {
+            j++;
+            next[i] = j;
+            i++;
+        } else if (j > 0) {
+            j = next[j - 1];
+        } else {
+            next[i] = 0;
+            i++;
+        }
+    }
+    return next;
+}
+
+bool repeatedSubstringPattern(char* s) {
+
+    int len = strlen(s);
+    int *next = build_next(s, len);
+    bool result = false;
+
+    // 检查最小重复片段能否被长度整除
+    if (next[len - 1]) {
+        result = len % (len - next[len - 1]) == 0;
+    }
+
+    free(next);
+    return result;
+}
+
+```
+
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">