Add Solution for 086 Partition List

cpp/086_Partition_List.cpp

@@ -0,0 +1,71 @@
+// 86. Partition List
+/**
+ * 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.
+ * 
+ * Tags: Linked List Two Pointers
+ * 
+ * Author: Kuang Qin
+ */
+
+#include <iostream>
+
+using namespace std;
+
+/**
+ * Definition for singly-linked list.
+ */
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode(int x) : val(x), next(NULL) {}
+    ListNode(int x, ListNode *p) : val(x), next(p) {}
+};
+
+class Solution {
+public:
+    ListNode* partition(ListNode* head, int x) {
+        if (head == NULL || head->next == NULL) {
+            return head;
+        }
+
+        // build two lists: l1 < x, l2 >= x
+        ListNode node1(0), node2(0);
+        ListNode *p1 = &node1, *p2 = &node2;
+
+        while (head != NULL) {
+            if (head->val < x) {
+                p1->next = head;
+                p1 = p1->next;
+            }
+            else {
+                p2->next = head;
+                p2 = p2->next;
+            }
+
+            head = head->next;
+        }
+
+        p2->next = NULL;        // the last element
+        p1->next = node2.next;  // connect two lists
+        return node1.next;
+    }
+};
+
+int main() {
+    ListNode node6(2), node5(5, &node6), node4(2, &node5), node3(3, &node4), node2(4, &node3), node1(1, &node2);
+    Solution sol;
+    ListNode *p = sol.partition(&node1, 3);
+    while (p != NULL) {
+        cout << p->val << " ";
+        p = p->next;
+    }
+    cin.get();
+
+    return 0;
+}