Note: The returned array must be malloced, assume caller calls free()
时间: 2024-05-15 18:15:19 浏览: 308
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.
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