leetcode-203-Remove Linked List Elements

                                Remove Linked List Elements

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

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

注意第一个节点和最后一个节点的处理

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        if (!head) return NULL;
        ListNode* pre,*h=head;
        while (h && h->val == val && h->next){  // 删除在前面所有值为val的节点
            h->val = h->next->val ;
            h->next = h->next->next ;
        }
        if (h->val == val) return NULL; //  此时 必定只有一个元素
        else {   // pre 为 h 前面的节点
            pre = h;
            h = h->next;
        }
        while (h) { // 删除值为val的节点
            if (h->val == val) {
                pre->next = h->next;
                h = h->next;
            }
            else {
                pre = h;
                h = h->next;
            }
        }
        return head; // 返回头结点
    }
};

在head前面加一个节点，便于处理，这样就可以不考虑首尾节点的处理过程。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* newH;
        newH->next = head;
        ListNode* cur = newH;
        while (cur->next) {
            if (cur->next->val == val) {
                cur->next = cur->next->next;
            }
            else cur = cur->next; 
        }
        return newH->next;
    }
};

leetcode-203-Remove Linked List Elements

原文：http://blog.csdn.net/u014705854/article/details/46916177