国产xxxx99真实实拍_久久不雅视频_高清韩国a级特黄毛片_嗯老师别我我受不了了小说

資訊專欄INFORMATION COLUMN

[LintCode/LeetCode] Partition List

崔曉明 / 3546人閱讀

摘要：新建兩個鏈表，分別存和的結點。令頭結點分別叫作和，對應的指針分別叫作和。然后遍歷，當小于的時候放入，否則放入。最后，讓較小值鏈表尾結點指向較大值鏈表頭結點，再讓較大值鏈表尾結點指向。

Problem

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.

Note

新建兩個鏈表，分別存和>=x的結點。令頭結點分別叫作dummyleft和dummyright，對應的指針分別叫作left和right。然后遍歷head，當head.val小于x的時候放入left.next，否則放入right.next。最后，讓較小值鏈表尾結點left指向較大值鏈表頭結點dummyright.next，再讓較大值鏈表尾結點right指向null。這樣兩個新鏈表就相連了，返回頭結點dummyleft.next即可。

Solution 
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode dummyleft = new ListNode(0);
        ListNode dummyright = new ListNode(0);
        ListNode left = dummyleft, right = dummyright;
        while (head != null) {
            if (head.val < x) {
                left.next = head;
                left = left.next;
            }
            else {
                right.next = head;
                right = right.next;
            }
            head = head.next;
        }
        left.next = dummyright.next;
        right.next = null;
        return dummyleft.next;
    }
}

文章版權歸作者所有，未經允許請勿轉載,若此文章存在違規行為，您可以聯系管理員刪除。

轉載請注明本文地址：http://m.specialneedsforspecialkids.com/yun/66204.html

相關文章

  • [LintCode/LeetCode] Meeting Rooms

    Problem Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings. Example Given intervals = [[0,30],[...

    Noodles 評論0 收藏0

  • [LintCode/LeetCode] Nth to Last Node in List

    摘要：依然是一道找倒數第個結點的鏈表題，用雙指針做。先走，然后和一起走，直到為，的位置就是倒數第個位置。 Problem Find the nth to last element of a singly linked list. The minimum number of nodes in list is n. Example Given a List 3->2->1->5->null ...

    Salamander 評論0 收藏0

  • [LintCode/LeetCode] Copy List with Random Pointer

    摘要：大體意思就是，先復制到，順便將所有的放在再復制所有的到，順便將所有的放在最后令，令，將和分離，返回的頭結點 Problem A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. ...

    Jacendfeng 評論0 收藏0

  • [LintCode/LeetCode] Rotate List

    摘要：而后吾當依除取余之法，化大為小，則指針不致于越界也。后欲尋右起第結點，令快指針先行數日，及至兩指針相距為，便吟鞭東指，與慢指針策馬共進。快慢指針亦止于其所焉。舞動長劍，中宮直入，直取首級，而一掌劈空，已鴻飛冥冥。自此，一代天驕，霸業已成。 Problem Given a list, rotate the list to the right by k places, where k is...

    Blackjun 評論0 收藏0

  • [LintCode/LeetCode] Convert Sorted List to Balance

    摘要：當鏈表為空時，中出現大于，返回。然后計算中點，以為界分別遞歸構建左右子樹。順序是，左子樹根結點右子樹。由于根節點是直接取構建，當前的已經被取用。所以在下一次遞歸構建右子樹之前，要讓指向。最后將和左右子樹相連，返回。 Problem Given a singly linked list where elements are sorted in ascending order, conve...

    Michael_Ding 評論0 收藏0

發表評論

0條評論

最新活動
閱讀需要支付1元查看
<