《剑指offer》(第二版)Java实现
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License: MIT License
《剑指offer》(第二版)Java实现
License: MIT License
/**
* 常规方法,从first开始找到要删除结点的前一个结点,时间复杂度为O(n)
*/
public void deleteNode_2(Node first, Node toBeDel) {
if (first == null || toBeDel == null) {
return;
}
// 要删除的就是链表头,它没有前一个结点
if (first == toBeDel) {
first = first.next;
} else {
Node cur = first;
// 找到被删除结点的前一个结点
while (cur.next != toBeDel) {
cur = cur.next;
}
// cur为toBeDel的前一个结点
cur.next = cur.next.next;
}
}
老哥你好,第 18 题删除链表的节点中,上述代码的 else
语句块的 while
循环中,找到被删除结点的前一个结点这个我觉得可能会出现问题,没有中断条件的话,如果要删除的节点不在链表中会一直循环。我觉得应该要加一个中断条件,如果要删除节点不在链表中,并且已经遍历完了链表要退出循环。修改如下:
/**
* 常规方法,从first开始找到要删除结点的前一个结点,时间复杂度为O(n)
*/
public void deleteNode_2(Node first, Node toBeDel) {
if (first == null || toBeDel == null) {
return;
}
// 要删除的就是链表头,它没有前一个结点
if (first == toBeDel) {
first = first.next;
} else {
Node cur = first;
// 找到被删除结点的前一个结点
while (curr.next != null && cur.next != toBeDel) {
cur = cur.next;
}
if (curr.next == null) {
return;
}
// cur为toBeDel的前一个结点
cur.next = cur.next.next;
}
}
希望老哥看下这样可不可行,谢谢。
输入队列 3, 2, 1
此时的data: 3, 2, 1
此时的max: 3, 3, 3
把3出队后,max的返回值仍是3
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