LeetCode725. Split Linked List in Parts

链表拆分

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错误提示：

member access within misaligned address 0x000100000001 for type 'struct ListNode', which requires 8

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问题代码：

struct ListNode** lstArr = (struct ListNode**)malloc(k * sizeof(struct ListNode*) );   

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失败了

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
struct ListNode** splitListToParts(struct ListNode* root, int k, int* returnSize) {
    // 1. handle null case
    if(NULL == root) return NULL;

    // 2. allocate memory
    struct ListNode* lstArr = (struct ListNode*)malloc(k * sizeof(struct ListNode));
    returnSize = (int *)malloc(k * sizeof(int));
    for(int i=0; i<k; i++){
        lstArr[i].next = NULL;
    }



    // 3. compute length of list, && splitted list
    int nodeLen = 0, minimum = 0, rem = 0; 
    struct ListNode* pNode = root, *tail = NULL;
    while(NULL != pNode)  {
        nodeLen ++;
        pNode = pNode->next;
    }
    minimum = nodeLen/k, rem = nodeLen%k;
    //printf("%d\n", rem);
    for(int i=0; i<k; i++){
        returnSize[i] = minimum;
    }
    for(int j=0; j<rem; j++){
        returnSize[j] += 1;
    }

    // 4. split the list
    pNode = root;
    for(int i=0; i<k; i++){
        lstArr[i].val = pNode->val;
        if(NULL != pNode)
            lstArr[i].next = pNode->next;
        int t = returnSize[i];
        tail = NULL;
        while(NULL != pNode && t>0){
            tail = pNode;
            pNode = pNode->next;
            t--;
        }
        if(NULL != tail) tail->next = NULL;
    }


    // 5. return result
    return &lstArr;
}