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Update 0019.删除链表的倒数第N个节点.md

修改了0019.删除链表的倒数第N个节点 java 版本

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`‎problems/0019.删除链表的倒数第N个节点.md‎`

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Original file line number	Diff line number	Diff line change
`@@ -87,30 +87,31 @@ public:`
`87`	`87`	`java:`
`88`	`88`
`89`	`89`	```java
`90`		`-class Solution {`
`91`		`- public ListNode removeNthFromEnd(ListNode head, int n) {`
`92`		`- ListNode dummy = new ListNode(-1);`
`93`		`- dummy.next = head;`
`94`		`-`
`95`		`- ListNode slow = dummy;`
`96`		`- ListNode fast = dummy;`
`97`		`- while (n-- > 0) {`
`98`		`- fast = fast.next;`
`99`		`- }`
`100`		`- // 记住待删除节点slow 的上一节点`
`101`		`- ListNode prev = null;`
`102`		`- while (fast != null) {`
`103`		`- prev = slow;`
`104`		`- slow = slow.next;`
`105`		`- fast = fast.next;`
`106`		`- }`
`107`		`- // 上一节点的next指针绕过待删除节点slow 直接指向slow的下一节点`
`108`		`- prev.next = slow.next;`
`109`		`- // 释放待删除节点slow 的next指针, 这句删掉也能AC`
`110`		`- slow.next = null;`
	`90`	`+public ListNode removeNthFromEnd(ListNode head, int n){`
	`91`	`+ ListNode dummyNode = new ListNode(0);`
	`92`	`+ dummyNode.next = head;`
	`93`	`+`
	`94`	`+ //排除示例 2 的情况:head = [1], n = 1`
	`95`	`+ if (head == null)`
	`96`	`+ return null;`
`111`	`97`
`112`		`- return dummy.next;`
	`98`	`+ ListNode fastIndex = dummyNode;`
	`99`	`+ ListNode slowIndex = dummyNode;`
	`100`	`+`
	`101`	`+ //只要快慢指针相差 n 个结点即可`
	`102`	`+ for (int i = 0; i < n ; i++){`
	`103`	`+ fastIndex = fastIndex.next;`
	`104`	`+ }`
	`105`	`+`
	`106`	`+ while (fastIndex.next != null){`
	`107`	`+ fastIndex = fastIndex.next;`
	`108`	`+ slowIndex = slowIndex.next;`
`113`	`109`	`}`
	`110`	`+`
	`111`	`+ //此时 slowIndex 的位置就是待删除元素的前一个位置。`
	`112`	`+ //具体情况可自己画一个链表长度为 3 的图来模拟代码来理解`
	`113`	`+ slowIndex.next = slowIndex.next.next;`
	`114`	`+ return dummyNode.next;`
`114`	`115`	`}`
`115`	`116`	```
`116`	`117`

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`‎problems/0019.删除链表的倒数第N个节点.md‎`

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