LeetCode86. 分隔链表

LeetCode86. 分隔链表

1. 问题描述

2. 思路

双指针

1. 声明两个节点small和big
2. 遍历当前链表
   • 如果cur.Val < x，将cur节点连接到small链表上
   • 否则，将cur节点连接到big链表上
3. 将small链表和big链表连接起来

3. 代码

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func partition(head *ListNode, x int) *ListNode {
    small, big := &ListNode{}, &ListNode{}
    cur,smallHead,bigHead := head, small,big

    for cur != nil {
        if cur.Val < x {
            small.Next = cur 
            small = small.Next
        } else {
            big.Next = cur
            big = big.Next
        }
        cur = cur.Next
    }
    // 最后要给big.Next赋值
    big.Next = nil
    small.Next = bigHead.Next
    return smallHead.Next
}