首页int** returnColumnSizes = (int**)malloc(sizeof(int*) * 100);

int returnColumnSizes = (int)malloc(sizeof(int) 100);

时间: 2023-10-04 21:12:02 浏览: 110

将字节转换成单位大小

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将字节大小转换成单位大小当文件上传后将文件大小获取与此同时该文件为字节大小将其转换成带有单位大小

This line of code allocates memory for an array of 100 integer pointers. The purpose of this array is to store the size of each column in a 2D array. The syntax used here is known as dynamic memory allocation, which allows the program to request memory from the operating system at runtime. In this case, the size of the memory block is determined by the size of the data type (int*) multiplied by the number of elements (100). The resulting memory block is then cast to type int**, which is a pointer to a pointer of integers. This means that the array can be treated as a 2D array, with each element pointing to a row of integers. It is important to note that this memory block must be freed using the free() function when it is no longer needed, to avoid memory leaks in the program.

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int** ret; int retSize; int* retColSize; int* path; int pathSize; void dfs(struct TreeNode* root, int targetSum) { if (root == NULL) { return; } path[pathSize++] = root->val; targetSum -= root->val; if (root->left == NULL && root->right == NULL && targetSum == 0) { int* tmp = malloc(sizeof(int) * pathSize); memcpy(tmp, path, sizeof(int) * pathSize); ret[retSize] = tmp; retColSize[retSize++] = pathSize; } dfs(root->left, targetSum); dfs(root->right, targetSum); pathSize--; } int** pathSum(struct TreeNode* root, int targetSum, int* returnSize, int** returnColumnSizes) { ret = malloc(sizeof(int) 2001); retColSize = malloc(sizeof(int) * 2001); path = malloc(sizeof(int) * 2001); retSize = pathSize = 0; dfs(root, targetSum); returnColumnSizes = retColSize; returnSize = retSize; return ret; }这段代码的作用以及每个变量的作用以及含义

这段代码实现了求解二叉树中所有从根节点到叶子节点路径上节点值之和等于目标值的路径。...targetSum表示目标值，returnSize表示返回二维数组的行数，returnColumnSizes是一个一维数组，用于存储每个路径的列数。

#include＜stdio.h＞ #include＜stdlib.h＞ void main() { int a, b, c; a=b=c=(int )malloc(sizeof(int)); a=1; b=2; c=3; a=b; printf("%d%d%d \n", a, b, c); }

a = b = c = (int *)malloc(sizeof(int)); *a = 1; *b = 2; *c = 3; b = a; // 修改为 b = a; printf("%d %d %d\n", *a, *b, *c); free(a); // 释放动态分配的内存 return 0; } 修正后的代码输出为...

bool isHaveSame(int * nums,int indexleft,int indexRight) { while (indexleft < indexRight) { if (nums[indexleft] == nums[indexRight]) return true; indexRight--; } return false; } void swap(int * nums,int indexA,int indexB) { int temp = nums[indexA]; nums[indexA]= nums[indexB]; nums[indexB]= temp; } void prem(int* nums, int numsSize, int* returnSize, int returnColumnSizes,int returnNums,int offset) { if(offset == numsSize) { //遍历到末尾了 //申请returnNums returnNums[returnSize] = (int )malloc(sizeof(int ) * numsSize); //拷贝内容到returnNums memcpy(returnNums[returnSize],nums,sizeof(int) numsSize ); //记录当前拷贝内容的长度 (returnColumnSizes)[returnSize] = numsSize; returnSize = returnSize + 1; } else { //回溯算法的核心 int index = offset; int i; for(i = offset; i < numsSize; i++) { if(!isHaveSame(nums,i,numsSize-1) ) { swap(nums,i,offset);//i 和 offset 交换 prem(nums,numsSize,returnSize,returnColumnSizes,returnNums,offset+1); swap(nums,i,offset);//i 和 offset 交换 } } } } int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) { //重复的数字的全排序 //组合次数为 n！= n ( n - 1) ( n - 2) ...... 2 * 1 //这样的方法适合回溯的方法 //取值范围1 <= nums.length <= 8 = 8* 7* 6 * 5 * 4 * 3 2 1 = 40320中可能 int returnNums = (int )malloc(sizeof(int ) 40320); returnColumnSizes= (int )malloc(sizeof(int ) * 40320); *returnSize = 0; prem(nums,numsSize,returnSize,returnColumnSizes,returnNums,0); return returnNums; }补全main函数

int *returnColumnSizes; int **returnNums = permuteUnique(nums, numsSize, &returnSize, &returnColumnSizes); for(int i = 0; i ; i++){ for(int j = 0; j < returnColumnSizes[i]; j++){ printf("%d ", ...

int merge(int intervals, int intervalsSize, int* intervalsColSize, int* returnSize, int** returnColumnSizes){ int c=intervalsColSize; for(int i=0;i<intervalsSize;i++) { for(int j=i+1;j<intervalsSize;j++) { if(intervals[i][c]>=intervals[j][0]) { if(intervals[i][c]<intervals[j][c]) { intervals[i][c]=intervals[j][c]; } for(int k=j+1;k<intervalsSize;k++) { intervals[k-1][0]=intervals[k][0]; intervals[k-1][c]=intervals[k][c]; intervalsSize--; } } returnColumnSizes[i]=intervalsColSize; } returnSize=&intervalsSize; return intervals; }检错

*returnColumnSizes = (int*)malloc(intervalsSize * sizeof(int)); int idx = 0; for (int i = 0; i ; i++) { int start = intervals[i][0], end = intervals[i][c - 1]; for (int j = i + 1; j ; j++) { if ...

returnColumnSizes = (int**)malloc(sizeof(int) 100); int level = 0; returnColumnSizes[*(returnSize)][j] = level;为什么报内存不对齐

因此，应该将第一行代码修改为int* returnColumnSizes = (int*)malloc(sizeof(int) * 100)。 2. 在第三行代码中，returnColumnSizes[*(returnSize)][j] = level;可能会导致越界访问，因为returnSize和j的值未知，不...

int cmp(const void p1, const void p2) { int pp1 = (int**)p1; int pp2 = (int)p2; // 若身高相同，则按照k从小到大排列 // 若身高不同，按身高从大到小排列 return pp1[0] == pp2[0] ? pp1[1] - pp2[1] : pp2[0] - pp1[0]; } // 将start与end中间的元素都后移一位 // start为将要新插入元素的位置 void moveBack(int people, int peopleSize, int start, int end) { int i; for(i = end; i > start; i--) { people[i] = people[i-1]; } } int reconstructQueue(int people, int peopleSize, int* peopleColSize, int* returnSize, int** returnColumnSizes){ int i; // 将people按身高从大到小排列（若身高相同，按k从小到大排列） qsort(people, peopleSize, sizeof(int), cmp); for(i = 0; i < peopleSize; ++i) { // people[i]要插入的位置 int position = people[i][1]; int temp = people[i]; // 将position到i中间的元素后移一位 // 注：因为已经排好序，position不会比i大。(举例：排序后people最后一位元素最小，其可能的k最大值为peopleSize-2，小于此时的i) moveBack(people, peopleSize, position, i); // 将temp放置到position处 people[position] = temp; } // 设置返回二维数组的大小以及里面每个一维数组的长度 returnSize = peopleSize; returnColumnSizes = (int)malloc(sizeof(int) peopleSize); for(i = 0; i < peopleSize; ++i) { (*returnColumnSizes)[i] = 2; } return people; }

cmp 函数中，通过将 p1 和 p2 强制转换为 int** 类型，并取其指向的地址的值，得到了 pp1 和 pp2 分别指向 p1 和 p2 所指向的一维数组。然后按照身高从大到小排列，如果身高相同，则按照 k 值从小...

c语言定义一个动态大小的二维数组，不允许c99修改下列代码：int** arrays; arrays = (int)malloc(factorial(mustPass_num) * sizeof(int)); **arrays = (int)malloc(mustPass_num sizeof(mustPass[0]));

arrays = (int**)malloc(mustPass_num * sizeof(int*)); for (int i = 0; i ; i++) { arrays[i] = (int*)malloc(factorial(mustPass_num) * sizeof(int)); } 这里首先分配了一个指向指针的指针 arrays，然后...

int maze; char path; maze=(int**)malloc(sizeof(int*)); path=(char**)malloc(sizeof(char*)); 这样定义二维指针有问题吗

maze = (int**)malloc(row * sizeof(int*)); path = (char**)malloc(row * sizeof(char*)); for (int i = 0; i ; i++) { maze[i] = (int*)malloc(col * sizeof(int)); path[i] = (char*)malloc(col * sizeof(char)...

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int** returnColumnSizes = (int**)malloc(sizeof(int*) * 100);

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int returnColumnSizes = (int)malloc(sizeof(int) 100);

int** returnColumnSizes；returnColumnSizes=(int)malloc(sizeof(int)2009); (returnColumnSizes)[*returnSize] = size;指针详解

int returnColumnSizes = (int)malloc(sizeof(int) 100); is this correct

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int returnColumnSizes = (int)malloc(sizeof(int) 100); why should we cast the returned pointer to an int double pointer

int returnColumnSizes = (int)malloc(sizeof(int) * 2000); what is the problem

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