LeetCode-086-分隔链表

分隔链表

解法一：链表遍历

public class LeetCode_086 {
    public static ListNode partition(ListNode head, int x) {
        // 小于x的链表节点
        ListNode lessThan = new ListNode(-1);
        // 不小于x的链表节点
        ListNode moreThan = new ListNode(-1);
        ListNode curLess = lessThan, curMore = moreThan;
        while (head != null) {
            if (head.val < x) {
                // 小于x的节点添加到链表lessThan中
                curLess.next = head;
                curLess = curLess.next;
            } else {
                // 不小于x的节点添加到链表moreThan中
                curMore.next = head;
                curMore = curMore.next;
            }
            head = head.next;
        }
        // 所有节点遍历完成后将lessThan和moreThan尾结点指向null
        curLess.next = null;
        curMore.next = null;
        // 将小于x的节点挪到不小于x的节点的前面
        curLess.next = moreThan.next;
        return lessThan.next;
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        head.next = new ListNode(4);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(2);
        head.next.next.next.next = new ListNode(5);
        head.next.next.next.next.next = new ListNode(2);

        ListNode result = partition(head, 3);
        while (result != null) {
            System.out.print(result.val + " ");
            result = result.next;
        }
    }
}