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| 1 | +/* ============================================================================== |
| 2 | + * 功能描述：LinkesListBasic |
| 3 | + * 创 建 者：wack |
| 4 | + * 创建日期：2017/4/28 18:42:06 |
| 5 | + * ==============================================================================*/ |
| 6 | +using System; |
| 7 | +using System.Collections.Generic; |
| 8 | +using System.Linq; |
| 9 | +using System.Text; |
| 10 | + |
| 11 | +namespace LinkedList.Lib |
| 12 | +{ |
| 13 | + /// <summary> |
| 14 | + /// LinkesListBasic |
| 15 | + /// </summary> |
| 16 | + public class LinkesListBasic |
| 17 | + { |
| 18 | + #region 构造函数 |
| 19 | + |
| 20 | + public LinkesListBasic() |
| 21 | + { |
| 22 | + } |
| 23 | + |
| 24 | + #endregion |
| 25 | + |
| 26 | + #region 属性字段 |
| 27 | + |
| 28 | + #endregion |
| 29 | + |
| 30 | + #region 私有方法 |
| 31 | + |
| 32 | + #endregion |
| 33 | + |
| 34 | + #region 公有方法 |
| 35 | + //Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. |
| 36 | + //Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, |
| 37 | + //the linked list should become 1 -> 2 -> 4 after calling your function. |
| 38 | + public void DeleteNode(ListNode node) |
| 39 | + { |
| 40 | + if (node == null) |
| 41 | + return; |
| 42 | + node.val = node.next.val; |
| 43 | + node.next = node.next.next; |
| 44 | + } |
| 45 | + |
| 46 | + //Given a sorted linked list, delete all duplicates such that each element appear only once. |
| 47 | + //For example, |
| 48 | + //Given 1->1->2, return 1->2. |
| 49 | + //Given 1->1->2->3->3, return 1->2->3. |
| 50 | + public ListNode DeleteDuplicates(ListNode head) |
| 51 | + { |
| 52 | + ListNode diff = head; |
| 53 | + ListNode tmp = head; |
| 54 | + while (tmp != null && tmp.next != null) |
| 55 | + { |
| 56 | + while (tmp.next != null && tmp.next.val == tmp.val) |
| 57 | + { |
| 58 | + tmp = tmp.next; |
| 59 | + } |
| 60 | + diff.next = tmp.next;//找到一个新值 |
| 61 | + diff = diff.next;//迭代 |
| 62 | + tmp = diff; |
| 63 | + } |
| 64 | + |
| 65 | + return head; |
| 66 | + } |
| 67 | + |
| 68 | + // Write a program to find the node at which the intersection of two singly linked lists begins. |
| 69 | + //For example, the following two linked lists: |
| 70 | + //A: a1 → a2 |
| 71 | + // ↘ |
| 72 | + // c1 → c2 → c3 |
| 73 | + // ↗ |
| 74 | + //B: b1 → b2 → b3 |
| 75 | + //begin to intersect at node c1. |
| 76 | + public ListNode GetIntersectionNode(ListNode headA, ListNode headB) |
| 77 | + { |
| 78 | + if (headA == null || headB == null) |
| 79 | + return null; |
| 80 | + ListNode a = headA; |
| 81 | + ListNode b = headB; |
| 82 | + while (a != b) |
| 83 | + { |
| 84 | + if (a == null) |
| 85 | + a = headB; |
| 86 | + else |
| 87 | + { |
| 88 | + a = a.next; |
| 89 | + } |
| 90 | + if (b == null) |
| 91 | + b = headA; |
| 92 | + else |
| 93 | + { |
| 94 | + b = b.next; |
| 95 | + } |
| 96 | + |
| 97 | + } |
| 98 | + return a; |
| 99 | + } |
| 100 | + |
| 101 | + //Remove all elements from a linked list of integers that have value val. |
| 102 | + //Example |
| 103 | + //Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6 |
| 104 | + //Return: 1 --> 2 --> 3 --> 4 --> 5 |
| 105 | + public ListNode RemoveElements(ListNode head, int val) |
| 106 | + { |
| 107 | + if (head == null) |
| 108 | + return null; |
| 109 | + //检测头是否等于val |
| 110 | + if (head.val == val) |
| 111 | + head = head.next; |
| 112 | + ListNode tmp = head; |
| 113 | + while (tmp != null && tmp.next != null) |
| 114 | + { |
| 115 | + if (tmp.next.val == val) |
| 116 | + { |
| 117 | + deleteNode(tmp, tmp.next); |
| 118 | + } |
| 119 | + else |
| 120 | + tmp = tmp.next; |
| 121 | + } |
| 122 | + //检测头是否等于val |
| 123 | + if (head != null && head.val == val) |
| 124 | + head = head.next; |
| 125 | + return head; |
| 126 | + } |
| 127 | + |
| 128 | + //删除节点node，pre为node的前驱 |
| 129 | + private void deleteNode(ListNode pre, ListNode node) |
| 130 | + { |
| 131 | + ListNode tmp = node.next; |
| 132 | + pre.next = tmp; |
| 133 | + } |
| 134 | + |
| 135 | + |
| 136 | + //Reverse a singly linked list. |
| 137 | + public ListNode ReverseList(ListNode head) |
| 138 | + { |
| 139 | + if (head == null || head.next == null) |
| 140 | + return head; |
| 141 | + ListNode a = head; |
| 142 | + ListNode b = head.next; |
| 143 | + a.next = null; |
| 144 | + while (b != null) |
| 145 | + { |
| 146 | + ListNode tmp = b.next; //保存节点b后的所有节点顺序 |
| 147 | + b.next = a; //上步保存后，可以放心的将b的next域指向a，实现反转 |
| 148 | + a = b; //上步实现反转后，再赋值给a，这样a始终为反转链表的头节点 |
| 149 | + b = tmp;//上步后实现了反转，这步实现迭代，即让b再在原来的链表中保持前行。 |
| 150 | + } |
| 151 | + return a; |
| 152 | + } |
| 153 | + |
| 154 | + //Merge two sorted linked lists and return it as a new list. |
| 155 | + //The new list should be made by splicing together the nodes of the first two lists. |
| 156 | + public ListNode MergeTwoLists(ListNode l1, ListNode l2) |
| 157 | + { |
| 158 | + if (l1 == null) return l2; |
| 159 | + if (l2 == null) return l1; |
| 160 | + |
| 161 | + ListNode merge; |
| 162 | + //首先确定merge的头节点指向谁 |
| 163 | + if (l1.val < l2.val) |
| 164 | + { |
| 165 | + merge = l1; |
| 166 | + l1 = l1.next; |
| 167 | + } |
| 168 | + else |
| 169 | + { |
| 170 | + merge = l2; |
| 171 | + l2 = l2.next; |
| 172 | + } |
| 173 | + |
| 174 | + ListNode im = merge; //临时指针指向merge |
| 175 | + while (l1 != null) |
| 176 | + { |
| 177 | + if (l2 != null) |
| 178 | + { |
| 179 | + if (l1.val < l2.val) |
| 180 | + { |
| 181 | + im.next = l1; |
| 182 | + l1 = l1.next; |
| 183 | + } |
| 184 | + else |
| 185 | + { |
| 186 | + im.next = l2; |
| 187 | + l2 = l2.next; |
| 188 | + } |
| 189 | + im = im.next; |
| 190 | + } |
| 191 | + else //b首先等于null |
| 192 | + { |
| 193 | + im.next = l1; |
| 194 | + break; |
| 195 | + } |
| 196 | + } |
| 197 | + |
| 198 | + if (l2 != null) //a首先等于null |
| 199 | + im.next = l2; |
| 200 | + |
| 201 | + return merge; |
| 202 | + |
| 203 | + } |
| 204 | + |
| 205 | + #endregion |
| 206 | + |
| 207 | + } |
| 208 | +} |
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