add solution of problem 86: partition list

Author: ppulse

PartitionList86/README.md

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+Given a linked list and a value *x*, partition it such that all nodes less than *x* come before nodes greater than or equal to *x*.
+
+You should preserve the original relative order of the nodes in each of the two partitions.
+
+For example,
+Given `1->4->3->2->5->2` and *x* = 3,
+return `1->2->2->4->3->5`.

PartitionList86/Solution.java

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+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    public ListNode partition(ListNode head, int x) {
+        if (head == null) {
+            return null;
+        }
+        
+        ListNode ltHead = null;
+        ListNode ltNode = null;
+        ListNode geHead = null;
+        ListNode geNode = null;
+        ListNode currNode = head;
+        
+        while (currNode != null) {
+            if (currNode.val < x) {
+                if (ltHead == null) {
+                    ltHead = currNode;
+                    ltNode = currNode;
+                } else {
+                    ltNode.next = currNode;
+                    ltNode = ltNode.next;
+                }
+            } else {
+                if (geHead == null) {
+                    geHead = currNode;
+                    geNode = currNode;
+                } else {
+                    geNode.next = currNode;
+                    geNode = geNode.next;
+                }
+            }
+            
+            currNode = currNode.next;
+        }
+        
+        if (geNode != null) {
+            geNode.next = null;
+        }
+        
+        if (ltNode != null) {
+            ltNode.next = geHead;
+        } else {
+            ltHead = geHead;
+        }
+        
+        return ltHead;
+    }
+}