206反转链表解题记录

主要是需要三个指针参与，然后一些细节

func reverseList(head *ListNode) *ListNode {
    curr := head
    if curr != nil { // 可能传入空链表
        nextone := curr.Next
        curr.Next = nil
        for nextone != nil {
            tmp := nextone.Next
            nextone.Next = curr
            curr = nextone
            nextone = tmp
        }
    }
    return curr
}

然后可以稍微优化一下减少一个判断，就是前驱节点其实最初可以设置为nil

func reverseList(head *ListNode) *ListNode {
    var prev *ListNode  // 显式初始化prev为nil
    curr := head
    for curr != nil {
        nextone := curr.Next
        curr.Next = prev
        prev = curr
        curr = nextone
    }
    return prev
}