add t57 t56 t37

链表/t36_两个链表的第一个公共节点.java

@@ -0,0 +1,53 @@
+/**
+ * 输入两个链表，找出它们的第一个公共结点。
+ */
+
+/**
+ * 特殊值测试：有一个链表为空
+ * 特殊值测试：无公共链表
+ */
+
+public class Solution {
+    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
+ 		int len1 = getLength(pHead1);
+        int len2 = getLength(pHead2);
+        if (len1 == 0 || len2 == 0) {
+            return null;
+        }
+        //计算链表差值
+        int dif = len1 > len2 ? len1 - len2 : len2 - len1;
+        //长的链表先走dif步
+        if (len1 > len2) {
+            for (int i = 0; i < dif; i++) {
+                pHead1 = pHead1.next;
+            }
+        } else {
+            for (int i = 0; i < dif; i++) {
+                pHead2 = pHead2.next;
+            }
+        }
+
+        //一起往前走
+        while (pHead1 != null && pHead2 != null && pHead1 != pHead2) {
+            pHead1 = pHead1.next;
+            pHead2 = pHead2.next;
+        }
+
+        if (pHead1 == null || pHead2 == null) {
+            return null;
+        } else {
+            return pHead1;
+        }
+
+    }
+
+    //获取链表长度
+    int getLength(ListNode pHead) {
+        int len = 0;
+        while (pHead != null) {
+            len++;
+            pHead = pHead.next;
+        }
+        return len;
+    }
+}

链表/t56_链表中环的入口节点.java

@@ -0,0 +1,62 @@
+/**
+ *	一个链表中包含环，请找出该链表的环的入口结点。
+ */
+
+/**
+ * 特殊输入：1.无环 2.整个链表为一个环
+ */
+
+public class Solution {
+
+    public ListNode EntryNodeOfLoop(ListNode pHead)
+    {
+        if (pHead == null) 
+            return null;
+        if (pHead.next == pHead)
+            return pHead;
+        //寻找环中的一个节点
+        ListNode node = nodeInLoop(pHead);
+        if (node == null)
+            return null;
+
+        //计算环中元素个数
+        ListNode node1 = node.next;
+        int count = 1;
+        while (node1 != node) {
+            count++;
+            node1 = node1.next;
+        }
+        //寻找入口
+        ListNode ahead = pHead;
+        ListNode behind = pHead;
+  		for (int i = 0; i < count; i++) {
+            ahead = ahead.next;
+        }
+       	while (ahead != behind) {
+            ahead = ahead.next;
+            behind = behind.next;
+        }
+        return ahead;
+
+
+    }
+    //通过一快一慢两个指针寻找环中的一个节点，无环时返回null
+    public ListNode nodeInLoop(ListNode pHead) {
+        if (pHead == null || pHead.next == null) {
+            return null;
+        }
+        ListNode slow = pHead;
+        ListNode fast = pHead;
+        slow = slow.next;
+        fast = fast.next.next;
+        while (fast != null && fast.next != null && slow != fast) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        if (fast == slow) 
+            return fast;
+        else 
+            return null;
+
+    }
+}

链表/t57_删除链表中重复的节点.java

@@ -0,0 +1,64 @@
+在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针。 例如，链表1->2->3->3->4->4->5 处理后为 1->2->5
+
+/**
+* 测试用例: 1->1->2->3->3
+**/
+public class Solution {
+    public ListNode deleteDuplication(ListNode pHead)
+    {
+		if (pHead == null) {
+            return null;
+        }
+        ListNode pre = null;
+        ListNode cur = pHead;
+        while (cur != null) {
+            ListNode next = cur.next;
+            boolean needDelete = false;
+            //判读是否删除当前节点
+            if (next != null && cur.val == next.val)
+                needDelete = true;
+            if (!needDelete) {
+                pre = cur;
+                cur = next;
+            } else {
+                //cur后移，直到为null或指向不重复元素
+                while (cur.next != null && cur.val == next.val) {
+                    cur = next;
+                    next = cur.next;
+                }
+                cur = next;
+                //更新pre
+                if (pre == null) {
+                    pHead = cur;
+                } else {
+                    pre.next = cur;
+                }
+
+            }
+        }
+        return pHead;
+    }
+}
+
+
+//除去重复元素，仅保留一个重复元素版本
+public class Solution {
+    public ListNode deleteDuplication(ListNode pHead)
+    {
+		if (pHead == null) {
+            return null;
+        }
+        ListNode cur = pHead;
+        while (cur.next != null) {
+            while (cur.next != null) {
+                if (cur.val == cur.next.val) {
+                 	cur.next = cur.next.next;   
+                } else {
+                    break;
+                }
+            }
+            cur = cur.next;
+        }
+        return pHead;
+    }
+}