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[LintCode/LeetCode] Partition List

崔曉明 / 3538人閱讀

摘要：新建兩個鏈表，分別存和的結(jié)點(diǎn)。令頭結(jié)點(diǎn)分別叫作和，對應(yīng)的指針分別叫作和。然后遍歷，當(dāng)小于的時候放入，否則放入。最后，讓較小值鏈表尾結(jié)點(diǎn)指向較大值鏈表頭結(jié)點(diǎn)，再讓較大值鏈表尾結(jié)點(diǎn)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(jié)點(diǎn)。令頭結(jié)點(diǎn)分別叫作dummyleft和dummyright，對應(yīng)的指針分別叫作left和right。然后遍歷head，當(dāng)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(jié)點(diǎn)left指向較大值鏈表頭結(jié)點(diǎn)dummyright.next，再讓較大值鏈表尾結(jié)點(diǎn)right指向null。這樣兩個新鏈表就相連了，返回頭結(jié)點(diǎn)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;
    }
}

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