标签：
My_Single_LinkedList
分4个部分实现（CRUD - 增删改查）。
首先要有一个Node（节点类）
1 class Node { 2 public int val; 3 public Node next; 4 5 public Node(int val) { 6 this.val = val; 7 this.next = null; 8 } 9 }
instance variables and constructer - 成员变量和构造函数
1 public Node head; //头指针指向虚拟的node，真正的list是取head.next 2 public Node last; //尾指针指向最后一个元素 3 public int length; //list的长度，不是index。在insert的时候，判断insert的index 4 5 public My_Single_LinkedList() { 6 this.head = new Node(-1); 7 this.last = head; 8 this.length = 0; 9 }
1. 添加
添加部分有两大部分，一个是添加在头和尾，一个是添加在某个index的左和右。
添加的同时要保持length和last指针的更新
1) 添加在头和尾：O(1)的时间复杂度
1 public void addLast(int val) { 2 last.next = new Node(val); 3 last = last.next; 4 this.length++; 5 } 6 7 public void addFirst(int val) { 8 Node newNode = new Node(val); 9 newNode.next = head.next; 10 head.next = newNode; 11 this.length++; 12 }
2) 在某个index的左和右添加：O(n)时间复杂度
1 /** 2 * insert before index 3 * @param val 4 * @param index 5 */ 6 public void insertBefore(int val, int index) { 7 if (index <= 0) { //在0之前加，相当于addFirst 8 addFirst(val); 9 return; 10 } 11 if (index >= length) {//在超出length之前加，相当于addLast 12 addLast(val); 13 return; 14 } 15 // 普通情况，两个指针，一前一后 16 Node cur = head.next; 17 Node prev = head; 18 int counter = 0; 19 while (cur != null && counter < index) { 20 cur = cur.next; 21 prev = prev.next; 22 counter++; 23 } 24 Node newNode = new Node(val); 25 newNode.next = cur; 26 prev.next = newNode; 27 this.length++; 28 } 29 30 /** 31 * insert after index 32 * @param val 33 * @param index certain index you want to insert at 34 */ 35 public void insertAfter(int val, int index) { 36 if (index < 0) { 37 addFirst(val); 38 return; 39 } 40 if (index >= length - 1) { 41 addLast(val); 42 return; 43 } 44 Node cur = head.next; 45 int counter = 0; 46 while (cur != null && counter < index) { 47 cur = cur.next; 48 counter++; 49 } 50 Node newNode = new Node(val); 51 newNode.next = cur.next; 52 cur.next = newNode; 53 this.length++; 54 }
2. 更新：O(n)时间复杂度
1 public void updateAt(int val, int index) { 2 if (index < 0 || index > length - 1) { 3 return; 4 } 5 int counter = 0; 6 Node cur = head.next; 7 while (cur != null && counter < index) { 8 cur = cur.next; 9 counter++; 10 } 11 cur.val = val; 12 }
3. 删除：O(n)时间复杂度
删除的时候也要注意length和last的更新
1 public void deleteAt(int index) { 2 if (index < 0 || index > length - 1) { 3 return; 4 } 5 int counter = 0; 6 Node prev = head; 7 Node cur = head.next; 8 while (cur != null && counter < index) { 9 cur = cur.next; 10 prev = prev.next; 11 counter++; 12 } 13 prev.next = prev.next.next; 14 //更新length和last指针 15 length--; 16 while (prev.next != null) { 17 prev = prev.next; 18 } 19 this.last = prev; 20 }
4. 搜索：O(n)时间复杂度
1 /** 2 * 根据val搜索 3 * 返回第一个符合条件node的index 4 * 如果没有，返回-1 5 */ 6 public int getByValue(int val) { 7 Node cur = head.next; 8 int index = 0; 9 10 while (cur != null) { 11 if (cur.val == val) { 12 return index; 13 } 14 cur = cur.next; 15 index++; 16 } 17 return -1; 18 } 19 20 /** 21 * 根据index搜索 22 */ 23 public Node get(int index) { 24 if (index < 0 || index > length - 1) { 25 return null; 26 } 27 Node cur = head.next; 28 int counter = 0; 29 while (cur != null && counter < index) { 30 cur = cur.next; 31 counter++; 32 } 33 return cur; 34 }
完整代码：
1 /** 2 * Created by Mingxiao on 9/5/2016. 3 */ 4 public class My_Single_LinkedList { 5 6 class Node { 7 public int val; 8 public Node next; 9 10 public Node(int val) { 11 this.val = val; 12 this.next = null; 13 } 14 } 15 16 public Node head; //头指针指向虚拟的node，真正的list是取head.next 17 public Node last; //尾指针指向最后一个元素 18 public int length; //list的长度，不是index。在insert的时候，判断insert的index 19 20 public My_Single_LinkedList() { 21 this.head = new Node(-1); 22 this.last = head; 23 this.length = 0; 24 } 25 // =============== INSERT========================== 26 public void addLast(int val) { 27 last.next = new Node(val); 28 last = last.next; 29 this.length++; 30 } 31 32 public void addFirst(int val) { 33 Node newNode = new Node(val); 34 newNode.next = head.next; 35 head.next = newNode; 36 this.length++; 37 } 38 39 /** 40 * insert before index 41 * @param val 42 * @param index 43 */ 44 public void insertBefore(int val, int index) { 45 if (index <= 0) { //在0之前加，相当于addFirst 46 addFirst(val); 47 return; 48 } 49 if (index >= length) {//在超出length之前加，相当于addLast 50 addLast(val); 51 return; 52 } 53 // 普通情况，两个指针，一前一后 54 Node cur = head.next; 55 Node prev = head; 56 int counter = 0; 57 while (cur != null && counter < index) { 58 cur = cur.next; 59 prev = prev.next; 60 counter++; 61 } 62 Node newNode = new Node(val); 63 newNode.next = cur; 64 prev.next = newNode; 65 this.length++; 66 } 67 68 /** 69 * insert after index 70 * @param val 71 * @param index certain index you want to insert at 72 */ 73 public void insertAfter(int val, int index) { 74 if (index < 0) { 75 addFirst(val); 76 return; 77 } 78 if (index >= length - 1) { 79 addLast(val); 80 return; 81 } 82 Node cur = head.next; 83 int counter = 0; 84 while (cur != null && counter < index) { 85 cur = cur.next; 86 counter++; 87 } 88 Node newNode = new Node(val); 89 newNode.next = cur.next; 90 cur.next = newNode; 91 this.length++; 92 } 93 94 //===================update========================= 95 public void updateAt(int val, int index) { 96 if (index < 0 || index > length - 1) { 97 return; 98 } 99 int counter = 0; 100 Node cur = head.next; 101 while (cur != null && counter < index) { 102 cur = cur.next; 103 counter++; 104 } 105 cur.val = val; 106 } 107 //=====================delete======================= 108 public void deleteAt(int index) { 109 if (index < 0 || index > length - 1) { 110 return; 111 } 112 int counter = 0; 113 Node prev = head; 114 Node cur = head.next; 115 while (cur != null && counter < index) { 116 cur = cur.next; 117 prev = prev.next; 118 counter++; 119 } 120 prev.next = prev.next.next; 121 //更新length和last指针 122 length--; 123 while (prev.next != null) { 124 prev = prev.next; 125 } 126 this.last = prev; 127 } 128 //===================search================================ 129 130 /** 131 * 根据val搜索 132 * 返回第一个符合条件node的index 133 * 如果没有，返回-1 134 */ 135 public int getByValue(int val) { 136 Node cur = head.next; 137 int index = 0; 138 139 while (cur != null) { 140 if (cur.val == val) { 141 return index; 142 } 143 cur = cur.next; 144 index++; 145 } 146 return -1; 147 } 148 149 /** 150 * 根据index搜索 151 */ 152 public Node get(int index) { 153 if (index < 0 || index > length - 1) { 154 return null; 155 } 156 Node cur = head.next; 157 int counter = 0; 158 while (cur != null && counter < index) { 159 cur = cur.next; 160 counter++; 161 } 162 return cur; 163 } 164 165 public void print() { 166 if (head.next == null) { 167 System.out.println("null"); 168 return; 169 } 170 Node cur = head.next; 171 while (cur.next != null) { 172 System.out.print(cur.val + "->"); 173 cur = cur.next; 174 } 175 System.out.println(cur.val); 176 } 177 178 179 public static void main(String[] args) { 180 My_Single_LinkedList list = new My_Single_LinkedList(); 181 list.addLast(1); 182 list.addLast(2); 183 list.addLast(0); 184 list.insertAfter(5,0); 185 list.updateAt(9, 1); 186 list.deleteAt(2); 187 System.out.println(list.get(2).val); 188 list.print(); 189 } 190 }
标签：
原文地址：http://www.cnblogs.com/marshallguo/p/5844303.html
