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‎README.md

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|周日|周一|周二|周三|周四|周五|周六|
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|:-:|:-:|:-:|:-:|:-:|:-:|:-:|
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|||||||[152](https://mp.weixin.qq.com/s/FvA_wcLq4JU70nXK6lrFHA)|
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|[171](https://mp.weixin.qq.com/s/jC4BsJ5r6QixdnwTIchTQw)|[16](https://mp.weixin.qq.com/s/IBEJdeU6G-sBSMuv02LrHg)|[24](https://mp.weixin.qq.com/s/_bjKxIrmtwU0IPtyWtKd_g)|[2](https://mp.weixin.qq.com/s/8cMt_Yaeu6AT5jk3DhdhqA)|[3](https://mp.weixin.qq.com/s/O-2CdSJTFWodVZzGYAjGrA)|
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|[171](https://mp.weixin.qq.com/s/jC4BsJ5r6QixdnwTIchTQw)|[16](https://mp.weixin.qq.com/s/IBEJdeU6G-sBSMuv02LrHg)|[24](https://mp.weixin.qq.com/s/_bjKxIrmtwU0IPtyWtKd_g)|[2](https://mp.weixin.qq.com/s/8cMt_Yaeu6AT5jk3DhdhqA)|[3](https://mp.weixin.qq.com/s/O-2CdSJTFWodVZzGYAjGrA)|[7](https://mp.weixin.qq.com/s/wwn4KOz_VBA1l2q6CHPdtA)|[9](https://mp.weixin.qq.com/s/hoBYW2m7oQ2h0UfwVjT5Hw)|
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|[19](https://mp.weixin.qq.com/s/qsElRxU6qJML9KQ3nWw9Yg)|[35](https://mp.weixin.qq.com/s/UnuRk-TT133I2OMxuD1hNg)|
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## 精选题解
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</a>
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<a href="">
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<a href="https://mp.weixin.qq.com/s/wwn4KOz_VBA1l2q6CHPdtA">
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<img width="24%" height="24%" src="https://i.loli.net/2019/06/05/5cf7854b1d01876390.png"/>
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</a>
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<a href="https://mp.weixin.qq.com/s/_bjKxIrmtwU0IPtyWtKd_g">
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<img width="24%" height="24%" src="https://i.loli.net/2019/05/26/5cea098e6bb7383654.png"/>
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</a>
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<a href="https://mp.weixin.qq.com/s/qsElRxU6qJML9KQ3nWw9Yg">
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<img width="24%" height="24%" src="https://i.loli.net/2019/06/08/5cfb590b01fc369826.png"/>
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</a>
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<a href="https://mp.weixin.qq.com/s/UnuRk-TT133I2OMxuD1hNg">
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<img width="24%" height="24%" src="https://i.loli.net/2019/06/10/5cfdaa6db660258479.png"/>
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</a>
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## 在线阅读开发
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在线阅读基于VuePress

‎docs/algorithm/19.md

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# 画解算法：19. 删除链表的倒数第N个节点
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## 题目链接
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https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/
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## 题目描述
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给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。
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**示例：**
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```bash
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给定一个链表: 1->2->3->4->5, 和 n = 2.
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当删除了倒数第二个节点后，链表变为 1->2->3->5.
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```
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**说明：**
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给定的 `n` 保证是有效的。
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**进阶：**
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你能尝试使用一趟扫描实现吗？
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## 解题方案
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### 思路
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- 标签：链表
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- 整体思路是让前面的指针先移动`n`步，之后前后指针共同移动直到前面的指针到尾部为止
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- 首先设立预先指针pre，预先指针是一个小技巧，在[第2题](https://mp.weixin.qq.com/s/8cMt_Yaeu6AT5jk3DhdhqA)中进行了讲解
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- 设预先指针`pre`的下一个节点指向`head`，设前指针为`start`，后指针为`end`，二者都等于`pre`
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- `start`先向前移动n步
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- 之后`start`和`end`共同向前移动，此时二者的距离为`n`，当`start`到尾部时，`end`的位置恰好为倒数第`n`个节点
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- 因为要删除该节点，所以要移动到该节点的前一个才能删除，所以循环结束条件为`start.next != null`
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- 删除后返回`pre.next`，为什么不直接返回`head`呢，因为`head`有可能是被删掉的点
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- 时间复杂度：O(n)
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### 代码
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```java
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode removeNthFromEnd(ListNode head, int n) {
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ListNode pre = new ListNode(0);
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pre.next = head;
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ListNode start = pre, end = pre;
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while(n != 0) {
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start = start.next;
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n--;
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}
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while(start.next != null) {
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start = start.next;
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end = end.next;
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}
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end.next = end.next.next;
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return pre.next;
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}
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}
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```
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### 画解
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![fr&lt;x&gt;ame_00001.png](https://i.loli.net/2019/06/08/5cfb590548c8068296.png)
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![fr&lt;x&gt;ame_00002.png](https://i.loli.net/2019/06/08/5cfb59056322550694.png)
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![fr&lt;x&gt;ame_00003.png](https://i.loli.net/2019/06/08/5cfb590548d3412898.png)
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![fr&lt;x&gt;ame_00004.png](https://i.loli.net/2019/06/08/5cfb5cb3942f072681.png)
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![fr&lt;x&gt;ame_00005.png](https://i.loli.net/2019/06/08/5cfb590acf6ef19427.png)
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![fr&lt;x&gt;ame_00006.png](https://i.loli.net/2019/06/08/5cfb5907c207755226.png)
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![fr&lt;x&gt;ame_00007.png](https://i.loli.net/2019/06/08/5cfb5908005cc43738.png)
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![fr&lt;x&gt;ame_00008.png](https://i.loli.net/2019/06/08/5cfb590b01fc369826.png)
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<span style="display:block;text-align:center;">点击「<strong>阅读原文</strong>」在PC端评论打卡</span>
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<span style="display:block;text-align:center;">后台回复「<strong>算法</strong>」，加入天天算法群</span>
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<span style="display:block;text-align:center;">觉得算法直击灵魂，欢迎点击<strong>在看</strong>和<strong>转发</strong></span>
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![](https://i.loli.net/2019/05/20/5ce23b33cc01d73486.gif)

‎docs/algorithm/35.md

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# 画解算法：35. 搜索插入位置
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## 题目链接
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https://leetcode-cn.com/problems/search-insert-position/
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## 题目描述
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给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
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你可以假设数组中无重复元素。
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示例 1:
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```bash
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输入: [1,3,5,6], 5
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输出: 2
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```
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示例 2:
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```bash
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输入: [1,3,5,6], 2
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输出: 1
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```
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示例 3:
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```bash
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输入: [1,3,5,6], 7
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输出: 4
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```
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示例 4:
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```bash
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输入: [1,3,5,6], 0
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输出: 0
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```
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## 解题方案
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### 思路
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- 标签：二分查找
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- 如果该题目暴力解决的话需要O(n)的时间复杂度，但是如果二分的话则可以降低到O(logn)的时间复杂度
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- 整体思路和普通的二分查找几乎没有区别，先设定左侧下标`left`和右侧下标`right`，再计算中间下标`mid`
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- 每次根据`nums[mid]`和`target`之间的大小进行判断，相等则直接返回下标，`nums[mid]<target`则left右移，`nums[mid]>target`则right左移
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- 查找结束如果没有相等值则返回left，该值为插入位置
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- 时间复杂度：O(logn)
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二分查找的思路不难理解，但是边界条件容易出错，比如**循环结束条件中left和right的关系，更新left和right位置时要不要加1减1**。
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下面给出两个可以直接套用的模板，记住就好了，免除边界条件出错。
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```java
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class Solution {
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public int searchInsert(int[] nums, int target) {
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int left = 0, right = nums.length - 1; // 注意
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while(left <= right) { // 注意
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int mid = (left + right) / 2; // 注意
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if(nums[mid] == target) { // 注意
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// 相关逻辑
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} else if(nums[mid] < target) {
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left = mid + 1; // 注意
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} else {
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right = mid - 1; // 注意
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}
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}
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// 相关返回值
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return 0;
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}
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}
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```
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或
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```java
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class Solution {
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public int searchInsert(int[] nums, int target) {
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int left = 0, right = nums.length; // 注意
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while(left < right) { // 注意
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int mid = (left + right) / 2; // 注意
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if(nums[mid] == target) {
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// 相关逻辑
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} else if(nums[mid] < target) {
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left = mid + 1; // 注意
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} else {
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right = mid; // 注意
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}
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}
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// 相关返回值
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return 0;
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}
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}
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```
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### 代码
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```java
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class Solution {
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public int searchInsert(int[] nums, int target) {
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int left = 0, right = nums.length - 1;
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while(left <= right) {
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int mid = (left + right) / 2;
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if(nums[mid] == target) {
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return mid;
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} else if(nums[mid] < target) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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}
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```
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### 画解
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![fr&lt;x&gt;ame_00001.png](https://i.loli.net/2019/06/10/5cfdaa6db556988046.png)
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![fr&lt;x&gt;ame_00002.png](https://i.loli.net/2019/06/10/5cfdaa6db2b2a58383.png)
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![fr&lt;x&gt;ame_00003.png](https://i.loli.net/2019/06/10/5cfdaa6d4400865526.png)
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![fr&lt;x&gt;ame_00004.png](https://i.loli.net/2019/06/10/5cfdaa6d8405114145.png)
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![fr&lt;x&gt;ame_00005.png](https://i.loli.net/2019/06/10/5cfdaa6d3f50a11555.png)
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![fr&lt;x&gt;ame_00006.png](https://i.loli.net/2019/06/10/5cfdaa6db660258479.png)
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<span style="display:block;text-align:center;">点击「<strong>阅读原文</strong>」在PC端评论打卡</span>
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<span style="display:block;text-align:center;">后台回复「<strong>算法</strong>」，加入天天算法群</span>
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<span style="display:block;text-align:center;">觉得算法直击灵魂，欢迎点击<strong>在看</strong>和<strong>转发</strong></span>
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![](https://i.loli.net/2019/05/20/5ce23b33cc01d73486.gif)

‎docs/algorithm/record.md

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# 题目记录
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# 题目打卡记录
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## 2019.05
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[112](https://mp.weixin.qq.com/s/_NdAjbWlYDV7z7LvMpuyDA)|[938](https://mp.weixin.qq.com/s/ni6l_a1QMCi2sKuZDNoe4Q)|[96](https://mp.weixin.qq.com/s/DEdiaun-RVwRj-9zx9brjw)|[101](https://mp.weixin.qq.com/s/9xiQ8KmTLk_gxf7lsie9dA)|[162](https://mp.weixin.qq.com/s/5HQ5zVPYdUAE4eVJBFKIoQ)|[674](https://mp.weixin.qq.com/s/jSUM4rmYXatKfrjACEdwvw)|
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|[704](https://mp.weixin.qq.com/s/mnfLuH1nU8ghShgOhw-KBg)|[77](https://mp.weixin.qq.com/s/VOV5wMsfrLW21ZkqXSfeeA)|||[520](https://mp.weixin.qq.com/s/YJTFk4xEOM9cDM-GQq_EIQ)|[242](https://mp.weixin.qq.com/s/nvxS4p3lXvAL21dDbJo8QA)|[1](https://mp.weixin.qq.com/s/jC4BsJ5r6QixdnwTIchTQw)|
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[112](https://mp.weixin.qq.com/s/_NdAjbWlYDV7z7LvMpuyDA)|[938](https://mp.weixin.qq.com/s/ni6l_a1QMCi2sKuZDNoe4Q)|[96](https://mp.weixin.qq.com/s/DEdiaun-RVwRj-9zx9brjw)|[101](https://mp.weixin.qq.com/s/9xiQ8KmTLk_gxf7lsie9dA)|[162](https://mp.weixin.qq.com/s/5HQ5zVPYdUAE4eVJBFKIoQ)|[674](https://mp.weixin.qq.com/s/jSUM4rmYXatKfrjACEdwvw)|92|
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|[704](https://mp.weixin.qq.com/s/mnfLuH1nU8ghShgOhw-KBg)|[77](https://mp.weixin.qq.com/s/VOV5wMsfrLW21ZkqXSfeeA)|[1](https://mp.weixin.qq.com/s/jC4BsJ5r6QixdnwTIchTQw)|260|[520](https://mp.weixin.qq.com/s/YJTFk4xEOM9cDM-GQq_EIQ)|[242](https://mp.weixin.qq.com/s/nvxS4p3lXvAL21dDbJo8QA)||
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## 2019.06

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