1.reverse-nodes-in-k-group（k組翻轉鏈表）【hard】

給你一個鏈表以及一個k,將這個鏈表從頭指針開始每k個翻轉一下。鏈表元素個數不是k的倍數，最後剩余的不用翻轉。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param head a ListNode
     * 
 
@param k an integer
     * @return a ListNode
     */
    public ListNode reverseKGroup(ListNode head, int k) {
        // Write your code here
        if (head == null || k <= 1) {
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
 
while (head.next != null) {
            head = reverseNextK(head, k);
        }
        return dummy.next;
    }
    private ListNode reverseNextK(ListNode head, int k) {
        ListNode node = head;
        for (int i = 0; i < k; i++) {
            if (node.next == null) {
                return
 
 head.next;
            }
            node = node.next;
        }
        ListNode n1 = head.next;
        ListNode prev = head;
        ListNode curt = head.next;
        for (int i = 0; i < k; i++) {
            ListNode temp = curt.next;
            curt.next = prev;
            prev = curt;
            curt = temp;
        }
        n1.next = curt;
        head.next = prev;
        return n1;
    }
}

註意：n0->n1->n2->...->nk->nk+1若要翻轉n1->...->nk,則n0到nk節點都會變化。 手動創建dummy node(哨兵節點)，dummy.next總是head(頭結點)，最後返回dummy.next也即head節點。reverseNextK()函數分三步：1.檢查是否有足夠的k個節點可翻轉（如果沒有,返回head.next,因為此時的head節點實際已在上次操作中被翻轉）2.翻轉（ n1 = head.next; prev = head; curt = head.next;for (int i = 0; i < k; i++) { temp = curt.next; curt.next = prev; prev = curt; curt = temp;}）3.鏈接，以便繼續下次翻轉（n1.next = curt; head.next = prev; return n1;）

2.reverse-linked-list（翻轉鏈表）

翻轉一個鏈表。給出一個鏈表1->2->3->null，這個翻轉後的鏈表為3->2->1->null。

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The head of linked list.
     * @return: The new head of reversed linked list.
     */
    public ListNode reverse(ListNode head) {
        // write your code here
        ListNode prev = null;
        ListNode curt = head;
        while (curt != null) {
            ListNode temp = curt.next;
            curt.next = prev;
            prev = curt;
            curt = temp;
        }
        return prev;
    }
}

3.partition-list（鏈表劃分）

給定一個單鏈表和數值x，劃分鏈表使得所有小於x的節點排在大於等於x的節點之前。你應該保留兩部分內鏈表節點原有的相對順序。

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param x: an integer
     * @return: a ListNode
     */
    public ListNode partition(ListNode head, int x) {
        // write your code here
        if (head == null) {
            return head;
        }
        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode left = leftDummy;
        ListNode right = rightDummy;
        while (head != null) {
            if (head.val < x) {
                left.next = head;
                left = head;
            } else {
                right.next = head;
                right = head;
            }
            head = head.next;
        }
        left.next = rightDummy.next;
        right.next = null;
        return leftDummy.next;
    }
}

註意：定義leftDummy和rightDummy兩個哨兵節點，分別用來保存<x和≥x節點，最後進行鏈接即可。在left(right).next=head之後，left(right)也要=head。

4.merge-two-sorted-lists（合並兩個排序鏈表）

將兩個排序鏈表合並為一個新的排序鏈表。

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        // write your code here
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                head.next = l1;
                l1 = l1.next;
            } else {
                head.next = l2;
                l2 = l2.next;
            }
            head = head.next;
        }
        if (l1 != null) {
            head.next = l1;
        } else {
            head.next = l2;
        }
        return dummy.next;
    }
}

註意：經典合並法。當兩個鏈表都不為空時，比較節點值的大小...當其中一個鏈表不為空時...

Chapter six Linked List & Array（鏈表與數組）