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Update 0019.删除链表的倒数第N个节点.md
原方法编译时发现对节点类型缺少断言。
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@ -188,18 +188,20 @@ TypeScript:
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```typescript
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function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
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let newHead: ListNode | null = new ListNode(0, head);
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let slowNode: ListNode | null = newHead,
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fastNode: ListNode | null = newHead;
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for (let i = 0; i < n; i++) {
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fastNode = fastNode.next;
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//根据leetcode题目的定义可推断这里快慢指针均不需要定义为ListNode | null。
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let slowNode: ListNode = newHead;
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let fastNode: ListNode = newHead;
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while(n--) {
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fastNode = fastNode.next!; //由虚拟头节点前进n个节点时,fastNode.next可推断不为null。
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}
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while (fastNode.next) {
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while(fastNode.next) { //遍历直至fastNode.next = null, 即尾部节点。 此时slowNode指向倒数第n个节点。
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fastNode = fastNode.next;
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slowNode = slowNode.next;
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slowNode = slowNode.next!;
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}
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slowNode.next = slowNode.next.next;
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return newHead.next;
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};
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slowNode.next = slowNode.next!.next; //倒数第n个节点可推断其next节点不为空。
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return newHead.next;
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}
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```
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版本二(计算节点总数法):
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