本周做一个单链表操作的题目。边写代码边测试，弄了近一个小时，中间有被中断，虽然一次就是acc了，但速度还是不够快。而且运行时间也不够快，分析了一下，可以使用头尾指针和长度运行时间会更快。比如插表尾法可以从O(n)降为O(1)。

这次给一个粗糙的算法。

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public class MyLinkedList {
	
	private class ListNode {
		
		 public int val;
		 
	     public ListNode next;
	     
	     public ListNode(int x) {
	    	 val = x; 
	    	 next = null;
	    }
	}
	
	private ListNode myLinkedList;
	
    /** Initialize your data structure here. */
    public MyLinkedList() {
    	myLinkedList = null;
    }
    
    /** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
    public int get(int index) {
    	if (null == myLinkedList) {
    		return -1;
    	}
    	int i = 0;
    	ListNode currNode = myLinkedList;
    	while (currNode != null) {
    		if (i == index) {
    			return currNode.val;
    		}
    		currNode = currNode.next;
    		++i;
    	}
        return -1;
    }
    
    /** Add a node of value val before the first element of the linked list. 
     *  After the insertion, the new node will be the first node of the linked list. 
     **/
    public void addAtHead(int val) {
        ListNode newNode = new ListNode(val);
        ListNode oldHeadNode = myLinkedList;
        newNode.next = oldHeadNode;
        myLinkedList = newNode;
    }
    
    /** Append a node of value val to the last element of the linked list. */
    public void addAtTail(int val) {
    	  ListNode newNode = new ListNode(val);
    	  // first node 
    	  if (null == myLinkedList) {
    		  myLinkedList = newNode;
    	  }
    	  ListNode preNode = null;
    	  ListNode currNode = myLinkedList;
    	  while (currNode != null) {
    		  preNode = currNode;
    		  currNode = currNode.next;
    	  }
    	  preNode.next = newNode;
    }
    
    /** Add a node of value val before the index-th node in the linked list.
     *  If index equals to the length of linked list, the node will be appended to the end of linked list.
     *  If index is greater than the length, the node will not be inserted. */
    public void addAtIndex(int index, int val) {
        if (index < 0) {
        	return;
        }
        if (0 == index) {
        	this.addAtHead(val);
        	return;
        }
        // linked list is empty, but index is greater than the length
        if (null == myLinkedList) {
        	return;
        }
        ListNode preNode = null;
        ListNode currNode = myLinkedList;
        int i = 0;
        while (currNode != null) {
        	if (i == index) {
            	break;
        	}
        	preNode = currNode;
        	currNode = currNode.next;
        	++i;
        }
        
        // when the index in linked list, then i == index, other is the length 
        if (index ==  i) {
        	ListNode newNode = new ListNode(val);
        	newNode.next = preNode.next;
        	preNode.next = newNode;
        }
        return;
    }
    
    /** Delete the index-th node in the linked list, if the index is valid. */
    public void deleteAtIndex(int index) {
        if (index < 0) {
        	return;
        }
        if (null == myLinkedList) {
        	return;
        }
        ListNode preNode = null;
        ListNode currNode = myLinkedList;
       
        while (currNode != null) {
        	if (0 == index) {  
        		if (preNode != null) { // find index node 
        			preNode.next = currNode.next;
        		} else { // first node
        			myLinkedList = currNode.next;
        		}
        		currNode = null;
        		break;
        	}
        	preNode = currNode;
        	currNode = currNode.next;
        	--index;
        }
    }
}