鏈表練習（一）

刪除鏈表中指定的所有元素

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if(head == null){
            return null;
        }
        ListNode prev = head;
        ListNode node = head.next;
        while(node != null){
            if(node.val == val){
                prev.next = node.next;
                node = prev.next;
            }
            else{
                prev = node;
                node = node.next;
            }
        }
        if(head.val ==val){
            head = head.next;
        }
        return head;
        }
}

反轉一個單鏈表

class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null){
            return null;
        }
        if(head.next == null){
            return head;
        }
        ListNode newHead = null;
        ListNode cur = head;
        ListNode prev = null;
        while(cur != null){
            ListNode next = cur.next;
            if(next == null){
                newHead = cur;
            }
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        return newHead;
    }
}

給定一個帶有頭結點 head 的非空單鏈表，返回鏈表的中間結點。

class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode cur = head;
        int steps = size(head)/2;
        for (int i = 0; i < steps; i++){
            cur = cur.next;
        }
        return cur;
    }
    int size(ListNode head){
        int size = 0;

            for( ; head!=null;head = head.next, size++);
        return size;
    }
}

輸出該鏈表中倒數第k個結點

public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        int len = size(head);
        if( k >len || k <= 0 || head == null){
            return null;
        } 
        int offset = len - k;
        ListNode cur = head;
        for(int i = 0; i < offset; i++){
            cur = cur.next;
        }
        return cur;
    }
    public int size(ListNode head){
        int size = 0;
        for(ListNode cur = head; cur != null; cur = cur.next){
            size++;
        }
        return size;
    }
}

將兩個有序鏈表合併爲一個新的有序鏈表並返回

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        ListNode newList = new ListNode(0);
        ListNode tail = newList;
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                tail.next = new ListNode(l1.val);
                l1 = l1.next;
                tail = tail.next;
            }
            else{
                tail.next = new ListNode(l2.val);
                l2 = l2.next;
                tail = tail.next;
            }
        }
        if( l1 != null ){
            tail.next = l1;
        }
        if( l2 != null ){
            tail.next = l2;
        }
        return newList.next;
    }
}