Cracking the Coding Interview 第二章

2.2 链表中倒数第k个结点

输入一个链表，输出该链表中倒数第k个结点。

思路：快慢指针(error: 判断是否有可行解，否则返回null, while, if 后加空格)

 /*
 public class ListNode {
     int val;
     ListNode next = null;

     ListNode(int val) {
         this.val = val;
     }
 }*/
 public class Solution {
     public ListNode FindKthToTail(ListNode head, int k) {
         if (head == null){
             return null;
         }
         ListNode fast = head;
         ListNode ans = head;
         for (int i = 0; i < k; i++){
             if (fast == null){
                 return null;
             }
             fast = fast.next;
         }
         while (fast != null){
             fast = fast.next;
             ans = ans.next;
         }
         return ans;
     }
 }

2.3 删除当前节点

实现一个算法，删除单向链表中间的某个结点，假定你只能访问该结点。
给定带删除的节点，请执行删除操作，若该节点为尾节点，返回false，否则返回true

思路：将当前的val转为下一个的val，释放下一个即可。

 import java.util.*;

 /*
 public class ListNode {
     int val;
     ListNode next = null;

     ListNode(int val) {
         this.val = val;
     }
 }*/
 public class Remove {
     public boolean removeNode(ListNode pNode) {
         // write code here
         if (pNode == null || pNode.next == null){
             return false;
         }
         //ListNode next = pNode.next;
         pNode.val = pNode.next.val;
         pNode.next = pNode.next.next;
         return true;
     }
 }

2.4 链表分割（小于x的在一边。大于等于x的又在一边）

编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前
给定一个链表的头指针 ListNode* pHead，请返回重新排列后的链表的头指针。注意：分割以后保持原来的数据顺序不变。

思路：（1）在原链表上找到第一个大的，将之后的小的往前移（error：移完之后不能立即向前，不然会错过连着的两个这种情况）

 import java.util.*;

 /*
 public class ListNode {
     int val;
     ListNode next = null;

     ListNode(int val) {
         this.val = val;
     }
 }*/
 public class Partition {
     public ListNode partition(ListNode pHead, int x) {
         // write code here
         if (pHead == null){
             return null;
         }
         ListNode front = new ListNode(0);
         front.next = pHead;
         pHead = front;
         while (pHead.next != null && pHead.next.val < x){
             pHead = pHead.next;
         }
         if (pHead.next != null){
             ListNode greater = pHead.next;
             while (greater.next != null){
                 if (greater.next.val < x){
                     ListNode temp = greater.next;
                     greater.next = temp.next;
                     temp.next = pHead.next;
                     pHead.next = temp;
                     pHead = temp;
                 } else {
                  greater = greater.next;
                 }
             }
         }
         pHead = front.next;
         front = null;
         return pHead;
     }
 }

（2）新建两个链表small large，指向对应元素即可。时间复杂度更大。

 import java.util.*;

 /*
 public class ListNode {
     int val;
     ListNode next = null;

     ListNode(int val) {
         this.val = val;
     }
 }*/
 public class Partition {
     public ListNode partition(ListNode pHead, int x) {
         // write code here
         /*if (pHead == null){
             return null;
         }
         ListNode front = new ListNode(0);
         front.next = pHead;
         pHead = front;
         while (pHead.next != null && pHead.next.val < x){
             pHead = pHead.next;
         }
         if (pHead.next != null){
             ListNode greater = pHead.next;
             while (greater.next != null){
                 if (greater.next.val < x){
                     ListNode temp = greater.next;
                     greater.next = temp.next;
                     temp.next = pHead.next;
                     pHead.next = temp;
                     pHead = temp;
                 } else {
                  greater = greater.next;
                 }
             }
         }
         pHead = front.next;
         front = null;
         return pHead;*/
         ListNode small = new ListNode(0);
         ListNode large = new ListNode(0);
         ListNode tailSmall = small;
         ListNode tailLarge = large;
         while (pHead != null){
             if (pHead.val < x){
                 tailSmall.next = new ListNode(pHead.val);
                 tailSmall = tailSmall.next;
             } else {
                 tailLarge.next = new ListNode(pHead.val);
                 tailLarge = tailLarge.next;
             }
             pHead = pHead.next;
         }
         tailSmall.next = large.next;
         return small.next;
     }
 }