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【Lintcode】096.Partition List

时间:2017-05-13 10:01:56      阅读:122      评论:0      收藏:0      [点我收藏+]

标签:ati   return   art   should   int   log   ret   题目   else   

题目:

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.

Example

Given 1->4->3->2->5->2->null and x = 3,
return 1->2->2->4->3->5->null.

题解:

Solution 1 ()

class Solution {  
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode* left = new ListNode(-1);
        ListNode* right = new ListNode(-1);
        ListNode* l = left;
        ListNode* r = right;
        while (head) {
            if (head->val < x) {
                l->next = head;
                l = l->next;
            } else {
                r->next = head;
                r = r->next;
            }
            head = head->next;
        }
        l->next = right->next;
        r->next = nullptr;
        
        return left->next;
    }
};

 

【Lintcode】096.Partition List

标签:ati   return   art   should   int   log   ret   题目   else   

原文地址:http://www.cnblogs.com/Atanisi/p/6847993.html

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