/** * Note: The returned array must be malloced, assume caller calls free(). */ int* twoSum(int* nums, int numsSize, int target,int* returnSize){ int i,j; int *result=(int *)malloc(sizeof(int )*2); for(i=0;i<numsSize;i++){ for(j=i+1;j<numsSize;j++){ if(nums[i]+nums[j]==target){ result[0]=i; result[1]=j; *returnSize=2; return result; } } } *returnSize=0; return NULL; }

这是一个关于 C 语言的问题，它是一个函数，用于在给定的整数数组中查找两个数，使它们的和等于目标值。如果找到这样的两个数，函数将返回它们的下标，否则返回空指针。函数还需要返回结果数组的大小，这个大小将被调用者用于释放内存。

相关问题

Note: The returned array must be malloced, assume caller calls free()

on the result array.

To allocate memory dynamically in C, we use the malloc() function. The syntax for using malloc() is as follows:

ptr = (castType*) malloc(size);

Here, ptr is a pointer of type castType. The malloc() function allocates a block of size bytes of memory, and returns a pointer to the first byte of the block. We can then use this pointer to access the block of memory.

In the context of the given problem, we need to allocate memory for a new array of size n, where n is the length of the input array. We can use malloc() to do this as follows:

int* result = (int*) malloc(n * sizeof(int));

Here, we first cast the pointer returned by malloc() to an int pointer, as we are allocating memory for an array of integers. We then multiply the size of an int (in bytes) by n, to get the total number of bytes to allocate. Finally, we assign the pointer returned by malloc() to the result variable.

The complete code for the function is as follows:

int* runningSum(int* nums, int numsSize, int* returnSize){
    int* result = (int*) malloc(numsSize * sizeof(int));
    int sum = 0;
    for (int i = 0; i < numsSize; i++) {
        sum += nums[i];
        result[i] = sum;
    }
    *returnSize = numsSize;
    return result;
}

Here, we first allocate memory for the result array using malloc(). We then iterate through the input array, computing the running sum and storing it in the result array. Finally, we set the value of returnSize to numsSize (as the length of the result array is equal to the length of the input array), and return the result array.

/** * Note: The returned array must be malloced, assume caller calls free(). */ int* twoSum(int* nums, int numsSize, int target,int* returnSize){ int i=0,j=0; for(i=0;i<numsSize;i++){ if(nums[i]<target){ target=target-nums[i]; for(j=i+1;j<numsSize;j++){ if(nums[j]==target){ returnSize[0]=i; returnSize[1]=j; break; } } } } return returnSize; }

这是一段实现两数之和的 C 代码。输入一个整数数组 nums，数组长度为 numsSize，以及目标值 target。函数返回一个长度为 2 的整数数组，其中包含两个元素的索引，这两个元素相加等于目标值 target。

该函数的实现过程如下：

1. 遍历整个数组 nums。
2. 对于每个元素 nums[i]，计算目标值 target 减去该元素的差值（即 diff = target - nums[i]）。
3. 在剩下的元素中找到一个值等于 diff 的元素，它的索引为 j。
4. 返回数组 [i, j]。

需要注意的是，该函数中的 returnSize 是一个指针，用于返回数组的长度。在函数内部，分配了一个长度为 2 的整数数组，用于存储返回结果。最后将该数组的指针返回即可。