From 80e4f2ff5650b428f29b8f1cdda714e5ae6a5116 Mon Sep 17 00:00:00 2001 From: CyC2018 <1029579233@qq.com> Date: Sat, 31 Mar 2018 11:51:44 +0800 Subject: [PATCH] auto commit --- notes/剑指 offer 题解.md | 7 ++----- pics/8433fbb2-c35c-45ef-831d-e3ca42aebd51.png | Bin 0 -> 3072 bytes pics/db4921d4-184b-48ba-a3cf-1d1141e3ba2d.png | Bin 0 -> 4096 bytes 3 files changed, 2 insertions(+), 5 deletions(-) create mode 100644 pics/8433fbb2-c35c-45ef-831d-e3ca42aebd51.png create mode 100644 pics/db4921d4-184b-48ba-a3cf-1d1141e3ba2d.png diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md index 4568effd..a60915dc 100644 --- a/notes/剑指 offer 题解.md +++ b/notes/剑指 offer 题解.md @@ -801,7 +801,7 @@ private void printNumber(char[] number) { ② 否则,就需要先遍历链表,找到节点的前一个节点,然后让前一个节点指向 null,时间复杂度为 O(N)。 -

+

综上,如果进行 N 次操作,那么大约需要操作节点的次数为 N-1+N=2N-1,其中 N-1 表示 N-1 个不是尾节点的每个节点以 O(1) 的时间复杂度操作节点的总次数,N 表示 1 个为节点以 O(n) 的时间复杂度操作节点的总次数。(2N-1)/N \~ 2,因此该算法的平均时间复杂度为 O(1)。 @@ -826,10 +826,7 @@ public ListNode deleteNode(ListNode head, ListNode tobeDelete) { ## 题目描述 -```html -Input : 1->2->3->3->4->4->5 -Output : 1->2->5 -``` +

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