Merge pull request #2855 from c-qwer/master

添加0459重复的子字符串 C 版本
This commit is contained in:
程序员Carl
2025-01-08 10:27:47 +08:00
committed by GitHub
4 changed files with 228 additions and 0 deletions

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@ -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;
}
```
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@ -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);
}
```
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@ -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;
}
```
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@ -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;
}
```
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