[LeetCode]203 Remove Linked List Elements

原题链接

Question

Remove all elements from a linked list of integers that have value val.

Example

Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6

Return: 1 –> 2 –> 3 –> 4 –> 5

My Answer

struct ListNode* removeElements(struct ListNode* head, int val) {
    struct ListNode* cur = head;
    struct ListNode* pre = NULL;

    while(cur){
        if(cur->val == val){
            if(pre){
                pre->next = cur->next;
                free(cur);
                cur = pre->next;
            }else{ //the value of the first node is equal to the given val.
                head = cur->next;
                //free(cur);
                cur = head;
            }
        }else{
            pre = cur;
            cur = cur->next;
        }
    }
    return head;
}

the complete code

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

//definition for the singly-linked list.
struct ListNode{
    int val;
    struct ListNode *next;
};

void build(struct ListNode* head, int* nums, int numsSize){
    struct ListNode *p, *q;

    p = head;
    for(int i=0; i<numsSize; ++i){
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->next = NULL;
        q->val = nums[i];
        p->next = q;
        p = q;
    }
}

void print(const struct ListNode* head){
    struct ListNode* p = head->next;
    while(p != NULL){
        printf("%d ", p->val);
        p = p->next;
    }
    printf("\n");
}

struct ListNode* removeElements(struct ListNode* head, int val) {
    struct ListNode* cur = head;
    struct ListNode* pre = NULL;

    while(cur){
        if(cur->val == val){
            if(pre){
                pre->next = cur->next;
                free(cur);
                cur = pre->next;
            }else{ //the value of the first node is equal to the given val.
                head = cur->next;
                //free(cur);
                cur = head;
            }
        }else{
            pre = cur;
            cur = cur->next;
        }
    }
    return head;
}

int main()
{
    struct ListNode* L = (struct ListNode*)malloc(sizeof(struct ListNode));
    L->next = NULL;
    L->val = -1;

    int A[] = {1,1};
    //int A[] = {};
    build(L, A, 2);

    print(L);

    struct ListNode* head = removeElements(L, 1);

    print(head);

    return 0;
}