LintCode #452 删除链表中的元素

方法很笨拙，被链表给绕住了，抽空在整理一下。

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param head: a ListNode
     * @param val: An integer
     * @return: a ListNode
     */
    public ListNode removeElements(ListNode head, int val) {
        // write your code here

        List<Integer> arr = getIndex(head,val);
        return removeElements(head, val, arr);
    }

    public static ListNode removeElements(ListNode head, int val, List<Integer> indexList) {

        ListNode result = head;
        int i =0;
        for (Integer index : indexList) {
            result = deleteNode(result, index - i);
            i++;
        }

        return result;

    }

    public static ListNode deleteNode(ListNode head, int index){

        //删除链表中的第一个元素
        if(index==0){
            head=head.next;
            return head;
        }
        int i=1;
        ListNode preNode=head;
        ListNode curNode=preNode.next;
        while(curNode!=null){
            if(i==index){
                preNode.next=curNode.next;
                break;
            }
            preNode=curNode;
            curNode=curNode.next;
            i++;
        }
        return head;
    }

    public static List<Integer> getIndex(ListNode head, int val){

        List<Integer> arr = new ArrayList<Integer>();
        if(head == null){
            return arr;
        }
        int i = 0;

        if(head.val == val) {
            arr.add(i);
        }
        ListNode curNode = head.next;
        i = 1;
        while (curNode != null) {

            if(curNode.val == val) {
                arr.add(i);
            }
            curNode = curNode.next;

            i++;
        }        

        return arr;
    }
}

C# 写法

        public static LinkNode GetLinkNode(LinkNode head, int val)
        {
            LinkNode result = head;
            while (result.Val == val)
            {
                result = result.Next;
            }

            LinkNode preNode = result;
            LinkNode curNode = preNode.Next;

            while (curNode != null)
            {
                if (curNode.Val == val)
                {
                    preNode.Next = curNode.Next;
                }

                preNode = curNode;
                curNode = curNode.Next;
            }

            return result;
        }