begeekmyfriend / leetcode

LeetCode in pure C

LEETCODE SOLUTIONS

Easy and Understandable C/C++ Solutions of Some Leetcode Questions.

leetcode's Issues

小建议

能添加下简单的算法描述吗

小白关于代码格式的问题请教

如 102 中的 10 行和 16 行

在有参宏定义时 #define 后接空格数量，第 10 行为一个，换行接四个。第 16 行为八个，换行接八个。

这样写是遵循了哪种规范？

0015 三数之和的函数定义处是 4 个参数，调用处是 3 个参数。

如题。

求教

int multiply(int a, char *b) {
return a b;
}

我就想问这道算法题如何解，codewars 的 c 注册入门题目

大神

我是C小白，不知道每个case要做什么😂

advice

Could you please give some annotations?Some codes are difficult to understand for me.

add_two_sum have something wrong

You convert it to reverse order before adding, but the carry should be given to the previous one.for example:243+563->342+365=707. in your code it's 608

Implement Non-Recursive Inorder Traversal for LeetCode Problem #94 to Avoid Stack Overflow

The current implementation of bst_inorder_traversal.c uses a recursive method for inorder traversal, which may lead to a stack overflow in cases of deep recursion. To mitigate this issue, it is recommended to use an iterative approach with an explicit stack. This prevents excessive stack usage and enhances the program's robustness. Below is the improved code:

#include <stdio.h>
#include <stdlib.h>

struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

struct Stack {
    struct TreeNode* node;
    struct Stack* next;
};

void push(struct Stack** stack, struct TreeNode* node) {
    struct Stack* new_node = (struct Stack*)malloc(sizeof(struct Stack));
    if (new_node == NULL) {
        exit(1);
    }
    new_node->node = node;
    new_node->next = *stack;
    *stack = new_node;
}

struct TreeNode* pop(struct Stack** stack) {
    if (*stack == NULL) {
        return NULL;
    }
    struct Stack* top = *stack;
    struct TreeNode* node = top->node;
    *stack = top->next;
    free(top);
    return node;
}

int isEmpty(struct Stack* stack) {
    return stack == NULL;
}

void inorderTraversal(struct TreeNode* root) {
    struct Stack* stack = NULL;
    struct TreeNode* current = root;

    while (current != NULL || !isEmpty(stack)) {
        while (current != NULL) {
            push(&stack, current);
            current = current->left;
        }
        current = pop(&stack);
        printf("%d ", current->val);
        current = current->right;
    }
}

This approach improves memory management and ensures the traversal can handle deeper trees without crashing.

014 longestCommonPrefix do not free memory