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	`1`	`+## 题目`
	`2`	`+`
	`3`	`+* 1669. 合并两个链表`
	`4`	`+`
	`5`	`+给你两个链表 list1 和 list2 ,它们包含的元素分别为 n 个和 m 个。`
	`6`	`+`
	`7`	`+请你将 list1 中下标从 a 到 b 的全部节点都删除,并将list2 接在被删除节点的位置。`
	`8`	`+`
	`9`	`+## 思路`
	`10`	`+`
	`11`	`+`
	`12`	`+## 代码`
	`13`	`+`
	`14`	+```php
	`15`	`+`
	`16`	`+class Solution {`
	`17`	`+`
	`18`	`+ /**`
	`19`	`+ * @param ListNode $list1`
	`20`	`+ * @param Integer $a`
	`21`	`+ * @param Integer $b`
	`22`	`+ * @param ListNode $list2`
	`23`	`+ * @return ListNode`
	`24`	`+ */`
	`25`	`+ function mergeInBetween($list1, $a, $b, $list2) {`
	`26`	`+ $n = 1;`
	`27`	`+ $left = $list1;`
	`28`	`+ $right = $list1;`
	`29`	`+ $dumyRight = null;`
	`30`	`+ $dumyRight->next = $right;`
	`31`	`+ while ($n <= $b) {`
	`32`	`+ $dumyRight->next = $right->next->next;`
	`33`	`+ if ($n < $a) {`
	`34`	`+ $left = $left->next;`
	`35`	`+ }`
	`36`	`+ $right = $right->next;`
	`37`	`+`
	`38`	`+ $n++;`
	`39`	`+ }`
	`40`	`+ $left->next = $list2;`
	`41`	`+ $cur = $list2;`
	`42`	`+ while ($cur->next) {`
	`43`	`+ $cur = $cur->next;`
	`44`	`+ }`
	`45`	`+ $cur->next = $dumyRight->next;`
	`46`	`+ return $list1;`
	`47`	`+ }`
	`48`	`+}`
	`49`	`+`
	`50`	+```

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