Given the head
of a singly linked list, return true
if it is a
or false
otherwise.
Example 1:
Input: head = [1,2,2,1] Output: true
Example 2:
Input: head = [1,2] Output: false
Constraints:
- The number of nodes in the list is in the range
[1, 105]
. 0 <= Node.val <= 9
Follow up: Could you do it in O(n)
time and O(1)
space?
A:
就是简单的读2遍,然后一个从后往前,一个从前往后。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */ class Solution { public: bool isPalindrome(ListNode* head) { vector<int> v; ListNode* runner = head; while(runner){ v.push_back(runner->val); runner = runner->next; } runner = head; while(runner){ int val = v.back(); v.pop_back(); if(val != runner->val){ return false; } runner = runner->next; } return true; } };
Follow up:
Could you do it in O(n)
time and O(1)
space?
1: reverse后半段。
2: 利用 recursive 的call stack,来对比
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