测试更改

Author: wangliang

.gitignore

@@ -1,3 +1,3 @@
-*~
-_book/
+*~
+_book/
 .DS_Store

README.md

@@ -1,21 +1,21 @@
-# 介绍
-
-题解预览地址：[https://leetcode.wang](https://leetcode.wang)，推荐电脑端打开，手机打开的话将页面滑到最上边，左上角是菜单
-
-leetcode 题目地址 https://leetcode.com/problemset/all/
-
-github 项目地址：https://github.com/wind-liang/leetcode 
-
-为什么刷题：https://leetcode.wang/leetcode100%E6%96%A9%E5%9B%9E%E9%A1%BE.html
-
-知乎开设了专栏，同步更新：[https://zhuanlan.zhihu.com/leetcode1024](https://zhuanlan.zhihu.com/leetcode1024)，关注后可以及时收到更新，网站有时候可能出问题打不开，建议关注一下知乎专栏备用
-
-准备刷一道，总结一道。
-
-可以加好友一起交流。
-
-微信： 17771420231
-
-公众号： windliang，更新编程相关
-
-如果觉得对你有帮助，记得给一个 star 哦 ^ ^
+# 介绍
+
+题解预览地址：[https://leetcode.wang](https://leetcode.wang)，推荐电脑端打开，手机打开的话将页面滑到最上边，左上角是菜单
+
+leetcode 题目地址 https://leetcode.com/problemset/all/
+
+github 项目地址：https://github.com/wind-liang/leetcode 
+
+为什么刷题：https://leetcode.wang/leetcode100%E6%96%A9%E5%9B%9E%E9%A1%BE.html
+
+知乎开设了专栏，同步更新：[https://zhuanlan.zhihu.com/leetcode1024](https://zhuanlan.zhihu.com/leetcode1024)，关注后可以及时收到更新，网站有时候可能出问题打不开，建议关注一下知乎专栏备用
+
+准备刷一道，总结一道。
+
+可以加好友一起交流。
+
+微信： 17771420231
+
+公众号： windliang，更新编程相关
+
+如果觉得对你有帮助，记得给一个 star 哦 ^ ^

SUMMARY.md

@@ -1,52 +1,52 @@
-{
-        "title": "leetcode",
-        "author": "windliang",
-        "description": "leetcode刷题",
-        "language" : "zh-hans", 
-        "links": {
-            "sidebar": {
-                "Home": "https://windliang.wang"
-            },
-            "gitbook": true
-        },
-        "structure": {
-            "readme": "README.md",
-            "summary": "SUMMARY.md"
-
-        }, 
-        "plugins": ["katex","splitter","anchor-navigation-ex","github","copy-code-button","-lunr", "-search", "search-plus","ad","-livereload","meta","sitemap","expandable-chapters-small"],
-        "pluginsConfig": {
-            "github": {
-                "url": "https://github.com/wind-liang/leetcode"
-            },
-            "anchor-navigation-ex": {
-                "multipleH1": false
-            },
-            "ad": {
-                "contentBottom": "%3Cdiv%20id%3D%22wechat_subscriber%22%20style%3D%22display%3A%20block%3B%20padding%3A%2010px%200%3B%20margin%3A%2020px%20auto%3B%20width%3A%20100%25%3B%20text-align%3A%20center%22%3E%0A%20%20%20%20%20%20%20%20%3Cimg%20id%3D%22wechat_subscriber_qcode%22%20src%3D%22https%3A//windliang.oss-cn-beijing.aliyuncs.com/leetcode.jpg%22%20alt%3D%22windliang%20wechat%22%20style%3D%22width%3A%20800px%3B%20max-width%3A%20100%25%3B%22%3E%0A%20%20%20%20%20%20%20%20%3Cbr%3E%20%0A%20%3Cbr%3E%20%0A%20%3Cbr%3E%20%0A%20%20%20%20%20%20%20%20%20%3Cdiv%3E%u5982%u679C%u89C9%u5F97%u6709%u5E2E%u52A9%u7684%u8BDD%uFF0C%u53EF%u4EE5%u70B9%u51FB%20%3Ca%20href%3D%22https%3A//github.com/wind-liang/leetcode%22%20target%20%3D%20%22_blank%22%3E%u8FD9%u91CC%3C/a%3E%20%u5728%20github%20%u7ED9%u4E00%u4E2A%20star%20%u54E6%20%5E%5E%3C/div%3E%0A%3Cbr%3E%0A%3Cdiv%3E%u5982%u679C%u60F3%u7CFB%u7EDF%u7684%u5B66%u4E60%u6570%u636E%u7ED3%u6784%u548C%u7B97%u6CD5%uFF0C%u5F3A%u70C8%u63A8%u8350%u4E00%u4E2A%u6211%u4E4B%u524D%u5B66%u8FC7%u7684%u8BFE%u7A0B%uFF0C%u53EF%u4EE5%u70B9%u51FB%20%3Ca%20href%3D%22http%3A//gk.link/a/105Cw%22%20target%20%3D%20%22_blank%22%3E%u8FD9%u91CC%3C/a%3E%20%u67E5%u770B%u8BE6%u60C5%3C/div%3E%0A%3Cbr%3E%0A%3Cdiv%3E%u8BFE%u7A0B%u5728%u516C%u4F17%u53F7%u300C%u8BFE%u7A0B%u51CF%u51CF%u300D%u8D2D%u4E70%u4F1A%20%3Ca%20href%3D%22https%3A//windliang.wang/2020/05/31/%25E6%259E%2581%25E5%25AE%25A2%25E6%2597%25B6%25E9%2597%25B4%25E4%25BC%2598%25E6%2583%25A0%25E7%25BA%25A2%25E5%258C%2585%25E8%25BF%2594%25E7%258E%25B0/%22%20target%20%3D%20%22_blank%22%3E%u66F4%u4F18%u60E0%u4E00%u4E9B%3C/a%3E%3C/div%3E%0A%20%20%20%20%20%20%20%20%3C/div%3E%3Cscript%20async%20src%3D%22https%3A//pagead2.googlesyndication.com/pagead/js/adsbygoogle.js%22%3E%3C/script%3E%3Cins%20class%3D%22adsbygoogle%22%20style%3D%22display%3Ablock%22%20data-ad-client%3D%22ca-pub-1480477141602191%22%20data-ad-slot%3D%228293832619%22%20data-ad-format%3D%22auto%22%20data-full-width-responsive%3D%22true%22%3E%3C/ins%3E%3Cscript%3E%20%28adsbygoogle%20%3D%20window.adsbygoogle%20%7C%7C%20%5B%5D%29.push%28%7B%7D%29%3B%3C/script%3E%0A%3Cdiv%20%20style%3D%22display%3A%20block%3B%20padding%3A%2010px%200%3B%20margin%3A%2020px%20auto%3B%20width%3A%20100%25%3B%20text-align%3A%20center%22%3E%0A%3Ca%20href%3D%22http%3A//www.beian.miit.gov.cn%22%20rel%3D%22nofollow%22%20%20target%3D%22_blank%22%20style%3D%22text-decoration%3Anone%3Bcolor%3A%23000%3B%22%3E%u664BICP%u590719005643%u53F7%3C/a%3E%0A%3C/div%3E"
-            },
-            "meta": {
-                "data": [
-                    {
-                        "name": "baidu-site-verification",
-                        "content": "iFzduP9Qsi"
-                    },
-                    {
-                        "name": "google-site-verification",
-                        "content": "qwgHd8kl7vEvBa4OR81jLieQG9UKcY2r8ocztMhzA8o"
-                    }
-                ]
-            },
-            "sitemap": {
-                "hostname": "https://leetcode.wang"
-            }
-            
-        },
-        "styles":{
-            "website":"websiteStyle.css"
-        },
-        "pdf": {
-            "fontSize": 16
-        },   
-        "gitbook" : "3.2.3"
+{
+        "title": "leetcode",
+        "author": "windliang",
+        "description": "leetcode刷题",
+        "language" : "zh-hans", 
+        "links": {
+            "sidebar": {
+                "Home": "https://windliang.wang"
+            },
+            "gitbook": true
+        },
+        "structure": {
+            "readme": "README.md",
+            "summary": "SUMMARY.md"
+
+        }, 
+        "plugins": ["katex","splitter","anchor-navigation-ex","github","copy-code-button","-lunr", "-search", "search-plus","ad","-livereload","meta","sitemap","expandable-chapters-small"],
+        "pluginsConfig": {
+            "github": {
+                "url": "https://github.com/wind-liang/leetcode"
+            },
+            "anchor-navigation-ex": {
+                "multipleH1": false
+            },
+            "ad": {
+                "contentBottom": "%3Cdiv%20id%3D%22wechat_subscriber%22%20style%3D%22display%3A%20block%3B%20padding%3A%2010px%200%3B%20margin%3A%2020px%20auto%3B%20width%3A%20100%25%3B%20text-align%3A%20center%22%3E%0A%20%20%20%20%20%20%20%20%3Cimg%20id%3D%22wechat_subscriber_qcode%22%20src%3D%22https%3A//windliang.oss-cn-beijing.aliyuncs.com/leetcode.jpg%22%20alt%3D%22windliang%20wechat%22%20style%3D%22width%3A%20800px%3B%20max-width%3A%20100%25%3B%22%3E%0A%20%20%20%20%20%20%20%20%3Cbr%3E%20%0A%20%3Cbr%3E%20%0A%20%3Cbr%3E%20%0A%20%20%20%20%20%20%20%20%20%3Cdiv%3E%u5982%u679C%u89C9%u5F97%u6709%u5E2E%u52A9%u7684%u8BDD%uFF0C%u53EF%u4EE5%u70B9%u51FB%20%3Ca%20href%3D%22https%3A//github.com/wind-liang/leetcode%22%20target%20%3D%20%22_blank%22%3E%u8FD9%u91CC%3C/a%3E%20%u5728%20github%20%u7ED9%u4E00%u4E2A%20star%20%u54E6%20%5E%5E%3C/div%3E%0A%3Cbr%3E%0A%3Cdiv%3E%u5982%u679C%u60F3%u7CFB%u7EDF%u7684%u5B66%u4E60%u6570%u636E%u7ED3%u6784%u548C%u7B97%u6CD5%uFF0C%u5F3A%u70C8%u63A8%u8350%u4E00%u4E2A%u6211%u4E4B%u524D%u5B66%u8FC7%u7684%u8BFE%u7A0B%uFF0C%u53EF%u4EE5%u70B9%u51FB%20%3Ca%20href%3D%22http%3A//gk.link/a/105Cw%22%20target%20%3D%20%22_blank%22%3E%u8FD9%u91CC%3C/a%3E%20%u67E5%u770B%u8BE6%u60C5%3C/div%3E%0A%3Cbr%3E%0A%3Cdiv%3E%u8BFE%u7A0B%u5728%u516C%u4F17%u53F7%u300C%u8BFE%u7A0B%u51CF%u51CF%u300D%u8D2D%u4E70%u4F1A%20%3Ca%20href%3D%22https%3A//windliang.wang/2020/05/31/%25E6%259E%2581%25E5%25AE%25A2%25E6%2597%25B6%25E9%2597%25B4%25E4%25BC%2598%25E6%2583%25A0%25E7%25BA%25A2%25E5%258C%2585%25E8%25BF%2594%25E7%258E%25B0/%22%20target%20%3D%20%22_blank%22%3E%u66F4%u4F18%u60E0%u4E00%u4E9B%3C/a%3E%3C/div%3E%0A%20%20%20%20%20%20%20%20%3C/div%3E%3Cscript%20async%20src%3D%22https%3A//pagead2.googlesyndication.com/pagead/js/adsbygoogle.js%22%3E%3C/script%3E%3Cins%20class%3D%22adsbygoogle%22%20style%3D%22display%3Ablock%22%20data-ad-client%3D%22ca-pub-1480477141602191%22%20data-ad-slot%3D%228293832619%22%20data-ad-format%3D%22auto%22%20data-full-width-responsive%3D%22true%22%3E%3C/ins%3E%3Cscript%3E%20%28adsbygoogle%20%3D%20window.adsbygoogle%20%7C%7C%20%5B%5D%29.push%28%7B%7D%29%3B%3C/script%3E%0A%3Cdiv%20%20style%3D%22display%3A%20block%3B%20padding%3A%2010px%200%3B%20margin%3A%2020px%20auto%3B%20width%3A%20100%25%3B%20text-align%3A%20center%22%3E%0A%3Ca%20href%3D%22http%3A//www.beian.miit.gov.cn%22%20rel%3D%22nofollow%22%20%20target%3D%22_blank%22%20style%3D%22text-decoration%3Anone%3Bcolor%3A%23000%3B%22%3E%u664BICP%u590719005643%u53F7%3C/a%3E%0A%3C/div%3E"
+            },
+            "meta": {
+                "data": [
+                    {
+                        "name": "baidu-site-verification",
+                        "content": "iFzduP9Qsi"
+                    },
+                    {
+                        "name": "google-site-verification",
+                        "content": "qwgHd8kl7vEvBa4OR81jLieQG9UKcY2r8ocztMhzA8o"
+                    }
+                ]
+            },
+            "sitemap": {
+                "hostname": "https://leetcode.wang"
+            }
+            
+        },
+        "styles":{
+            "website":"websiteStyle.css"
+        },
+        "pdf": {
+            "fontSize": 16
+        },   
+        "gitbook" : "3.2.3"
     }

book.json

@@ -1,101 +1,101 @@
-## 题目描述	（简单难度）
-
-![](http://windliang.oss-cn-beijing.aliyuncs.com/1_two_sum.jpg)
-
-给定一个数组和一个目标和，从数组中找两个数字相加等于目标和，输出这两个数字的下标。
-
-## 解法一
-
-简单粗暴些，两重循环，遍历所有情况看相加是否等于目标和，如果符合直接输出。
-
-``` JAVA
-public int[] twoSum(int[] nums, int target) {
-    int []ans=new int[2];
-    for(int i=0;i<nums.length;i++){
-        for(int j=(i+1);j<nums.length;j++){
-            if(nums[i]+nums[j]==target){
-                ans[0]=i;
-                ans[1]=j;
-                return ans;
-            }
-        }
-    }
-    return ans;
-}
-```
-
-
-时间复杂度：两层 for 循环，O（n²）
-
-空间复杂度：O（1）
-
-## 解法二
-
-在上边的解法中看下第二个 for 循环步骤。
-
-```JAVA
-for(int j=(i+1);j<nums.length;j++){
-	if(nums[i]+nums[j]==target){
-```
-
-我们换个理解方式：
-
-```JAVA
-for(int j=(i+1);j<nums.length;j++){ 
-	sub=target-nums[i]
- 	if(nums[j]==sub){
-```
-
-第二层 for 循环无非是遍历所有的元素，看哪个元素等于 sub ，时间复杂度为 O（n）。
-
-有没有一种方法，不用遍历就可以找到元素里有没有等于 sub 的？
-
-hash table ！！！
-
-我们可以把数组的每个元素保存为 hash 的 key，下标保存为 hash 的 value 。
-
-这样只需判断 sub 在不在 hash 的 key 里就可以了，而此时的时间复杂度仅为 O（1）！
-
-需要注意的地方是，还需判断找到的元素不是当前元素，因为题目里讲一个元素只能用一次。
-
-``` JAVA
-public int[] twoSum(int[] nums, int target) {
-    Map<Integer,Integer> map=new HashMap<>();
-    for(int i=0;i<nums.length;i++){
-        map.put(nums[i],i);
-    }
-    for(int i=0;i<nums.length;i++){
-        int sub=target-nums[i];
-        if(map.containsKey(sub)&&map.get(sub)!=i){
-            return new int[]{i,map.get(sub)};
-        }
-    }
-    throw new IllegalArgumentException("No two sum solution");
-}
-```
-
-时间复杂度：比解法一少了一个 for 循环，降为 O（n）
-
-空间复杂度：所谓的空间换时间，这里就能体现出来， 开辟了一个 hash table ，空间复杂度变为 O（n）
-
-## 解法三
-
-看解法二中，两个 for 循环，他们长的一样，我们当然可以把它合起来。复杂度上不会带来什么变化，变化仅仅是不需要判断是不是当前元素了，因为当前元素还没有添加进 hash 里。	
-
-``` JAVA
-public int[] twoSum(int[] nums, int target) {
-    Map<Integer,Integer> map=new HashMap<>();
-    for(int i=0;i<nums.length;i++){
-        int sub=target-nums[i];
-        if(map.containsKey(sub)){
-            return new int[]{i,map.get(sub)};
-        }
-        map.put(nums[i], i);
-    }
-    throw new IllegalArgumentException("No two sum solution");
-}
-```
-
-## 总结
-
+## 题目描述	（简单难度）
+
+![](http://windliang.oss-cn-beijing.aliyuncs.com/1_two_sum.jpg)
+
+给定一个数组和一个目标和，从数组中找两个数字相加等于目标和，输出这两个数字的下标。
+
+## 解法一
+
+简单粗暴些，两重循环，遍历所有情况看相加是否等于目标和，如果符合直接输出。
+
+``` JAVA
+public int[] twoSum(int[] nums, int target) {
+    int []ans=new int[2];
+    for(int i=0;i<nums.length;i++){
+        for(int j=(i+1);j<nums.length;j++){
+            if(nums[i]+nums[j]==target){
+                ans[0]=i;
+                ans[1]=j;
+                return ans;
+            }
+        }
+    }
+    return ans;
+}
+```
+
+
+时间复杂度：两层 for 循环，O（n²）
+
+空间复杂度：O（1）
+
+## 解法二
+
+在上边的解法中看下第二个 for 循环步骤。
+
+```JAVA
+for(int j=(i+1);j<nums.length;j++){
+	if(nums[i]+nums[j]==target){
+```
+
+我们换个理解方式：
+
+```JAVA
+for(int j=(i+1);j<nums.length;j++){ 
+	sub=target-nums[i]
+ 	if(nums[j]==sub){
+```
+
+第二层 for 循环无非是遍历所有的元素，看哪个元素等于 sub ，时间复杂度为 O（n）。
+
+有没有一种方法，不用遍历就可以找到元素里有没有等于 sub 的？
+
+hash table ！！！
+
+我们可以把数组的每个元素保存为 hash 的 key，下标保存为 hash 的 value 。
+
+这样只需判断 sub 在不在 hash 的 key 里就可以了，而此时的时间复杂度仅为 O（1）！
+
+需要注意的地方是，还需判断找到的元素不是当前元素，因为题目里讲一个元素只能用一次。
+
+``` JAVA
+public int[] twoSum(int[] nums, int target) {
+    Map<Integer,Integer> map=new HashMap<>();
+    for(int i=0;i<nums.length;i++){
+        map.put(nums[i],i);
+    }
+    for(int i=0;i<nums.length;i++){
+        int sub=target-nums[i];
+        if(map.containsKey(sub)&&map.get(sub)!=i){
+            return new int[]{i,map.get(sub)};
+        }
+    }
+    throw new IllegalArgumentException("No two sum solution");
+}
+```
+
+时间复杂度：比解法一少了一个 for 循环，降为 O（n）
+
+空间复杂度：所谓的空间换时间，这里就能体现出来， 开辟了一个 hash table ，空间复杂度变为 O（n）
+
+## 解法三
+
+看解法二中，两个 for 循环，他们长的一样，我们当然可以把它合起来。复杂度上不会带来什么变化，变化仅仅是不需要判断是不是当前元素了，因为当前元素还没有添加进 hash 里。	
+
+``` JAVA
+public int[] twoSum(int[] nums, int target) {
+    Map<Integer,Integer> map=new HashMap<>();
+    for(int i=0;i<nums.length;i++){
+        int sub=target-nums[i];
+        if(map.containsKey(sub)){
+            return new int[]{i,map.get(sub)};
+        }
+        map.put(nums[i], i);
+    }
+    throw new IllegalArgumentException("No two sum solution");
+}
+```
+
+## 总结
+
 题目比较简单，毕竟暴力的方法也可以解决。唯一闪亮的点就是，时间复杂度从 O（n²）降为 O（n） 的时候，对 hash 的应用，有眼前一亮的感觉。