Solve 234 with C++ and O(1) space!

Author: illuz

solutions/234.Palindrome_Linked_List/AC_reverse_1.cpp

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+/*
+*  Author:      illuz <iilluzen[at]gmail.com>
+*  File:        AC_reverse_1.cpp
+*  Create Date: 2015-07-29 11:49:16
+*  Descripton:   
+*/
+
+
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 0;
+
+// Definition for singly-linked list.
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode(int x) : val(x), next(NULL) {}
+};
+
+class Solution {
+public:
+    bool isPalindrome(ListNode* head) {
+        int sz = get_length(head);
+        if (sz < 2)
+            return true;
+
+        // reverse half
+        ListNode *new_head = reverseBetween(head, 1, sz / 2);
+
+        ListNode *head1 = new_head, *head2;
+        // now the head is point to (sz/2)th node
+        if (sz % 2 == 1) {
+            head2 = head->next->next;
+        } else {
+            head2 = head->next;
+        }
+        bool ok = true;
+        for (int i = 0; i < sz / 2; ++i) {
+            if (head1->val != head2->val) {
+                ok = false;
+                break;
+            }
+            head1 = head1->next;
+            head2 = head2->next;
+        }
+
+        // reverse half again
+        reverseBetween(new_head, 1, sz / 2);
+        return ok;
+    }
+
+private:
+    int get_length(ListNode* head) {
+        int sz = 0;
+        while (head) {
+            ++sz;
+            head = head->next;
+        }
+        return sz;
+    }
+
+    // these codes are the same as https://github.com/illuz/leetcode/blob/master/solutions/092.Reverse_Linked_List_II/AC_simulate_n.cpp
+    ListNode *reverseBetween(ListNode *head, int m, int n) {
+        ListNode *mhead = new ListNode(0), *prev, *cur;
+        mhead->next = head;    // because m will be 0
+        for (int i = 0; i < m - 1; i++) {
+            mhead = mhead->next;
+        }
+
+        prev = mhead->next;
+        cur = prev->next;
+        for (int i = m; i < n; i++) {
+            prev->next = cur->next;
+            cur->next = mhead->next;
+            mhead->next = cur;
+            cur = prev->next;
+        }
+        return m == 1 ? mhead->next : head;
+    }
+};
+
+int main() {
+    Solution s;
+    ListNode l1(1);
+    ListNode l2(2);
+    ListNode l3(2);
+    ListNode l4(1);
+    ListNode l5(2);
+    ListNode l6(1);
+    cout << s.isPalindrome(&l1) << endl;
+    l1.next = &l2;
+    cout << s.isPalindrome(&l1) << endl;
+    l2.next = &l3;
+    cout << s.isPalindrome(&l1) << endl;
+    l3.next = &l4;
+    cout << s.isPalindrome(&l1) << endl;
+    l4.next = &l5;
+    cout << s.isPalindrome(&l1) << endl;
+    l5.next = &l6;
+    cout << s.isPalindrome(&l1) << endl;
+    return 0;
+}
+