Merge remote-tracking branch 'origin/master'

# Conflicts:
#	src/main/java/leetcode/editor/cn/all.json

LICENSE

@@ -0,0 +1,201 @@
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src/main/java/leetcode/editor/cn/RegularExpressionMatching.java

@@ -0,0 +1,126 @@
+//给你一个字符串 s 和一个字符规律 p，请你来实现一个支持 '.' 和 '*' 的正则表达式匹配。 
+//
+// 
+// '.' 匹配任意单个字符 
+// '*' 匹配零个或多个前面的那一个元素 
+// 
+//
+// 所谓匹配，是要涵盖 整个 字符串 s的，而不是部分字符串。 
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：s = "aa" p = "a"
+//输出：false
+//解释："a" 无法匹配 "aa" 整个字符串。
+// 
+//
+// 示例 2: 
+//
+// 
+//输入：s = "aa" p = "a*"
+//输出：true
+//解释：因为 '*' 代表可以匹配零个或多个前面的那一个元素, 在这里前面的元素就是 'a'。因此，字符串 "aa" 可被视为 'a' 重复了一次。
+// 
+//
+// 示例 3： 
+//
+// 
+//输入：s = "ab" p = ".*"
+//输出：true
+//解释：".*" 表示可匹配零个或多个（'*'）任意字符（'.'）。
+// 
+//
+// 示例 4： 
+//
+// 
+//输入：s = "aab" p = "c*a*b"
+//输出：true
+//解释：因为 '*' 表示零个或多个，这里 'c' 为 0 个, 'a' 被重复一次。因此可以匹配字符串 "aab"。
+// 
+//
+// 示例 5： 
+//
+// 
+//输入：s = "mississippi" p = "mis*is*p*."
+//输出：false 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 0 <= s.length <= 20 
+// 0 <= p.length <= 30 
+// s 可能为空，且只包含从 a-z 的小写字母。 
+// p 可能为空，且只包含从 a-z 的小写字母，以及字符 . 和 *。 
+// 保证每次出现字符 * 时，前面都匹配到有效的字符 
+// 
+// Related Topics 字符串 动态规划 回溯算法 
+// 👍 2074 👎 0
+
+package leetcode.editor.cn;
+
+//10:正则表达式匹配
+public class RegularExpressionMatching {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new RegularExpressionMatching().new Solution();
+        //false
+        System.out.println(solution.isMatch("aa", "a"));
+        //true
+        System.out.println(solution.isMatch("aa", "a*"));
+        //true
+        System.out.println(solution.isMatch("aa", ".*"));
+        //true
+        System.out.println(solution.isMatch("ab", ".*"));
+        //false
+        System.out.println(solution.isMatch("aa", "a"));
+        //true
+        System.out.println(solution.isMatch("aab", "c*a*b"));
+        //false
+        System.out.println(solution.isMatch("aaa", "aaaa"));
+        //false
+        System.out.println(solution.isMatch("a", "ab*a"));
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public boolean isMatch(String s, String p) {
+            int sLength = s.length();
+            int pLength = p.length();
+
+            boolean[][] f = new boolean[sLength + 1][pLength + 1];
+            f[0][0] = true;
+            for (int i = 0; i <= sLength; ++i) {
+                for (int j = 1; j <= pLength; ++j) {
+                    if (p.charAt(j - 1) == '*') {
+                        f[i][j] = f[i][j - 2];
+                        if (matches(s, p, i, j - 1)) {
+                            f[i][j] = f[i][j] || f[i - 1][j];
+                        }
+                    } else {
+                        if (matches(s, p, i, j)) {
+                            f[i][j] = f[i - 1][j - 1];
+                        }
+                    }
+                }
+            }
+            return f[sLength][pLength];
+        }
+
+        public boolean matches(String s, String p, int i, int j) {
+            if (i == 0) {
+                return false;
+            }
+            if (p.charAt(j - 1) == '.') {
+                return true;
+            }
+            return s.charAt(i - 1) == p.charAt(j - 1);
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/RegularExpressionMatching.md

@@ -0,0 +1,60 @@
+<p>给你一个字符串 <code>s</code> 和一个字符规律 <code>p</code>，请你来实现一个支持 <code>'.'</code> 和 <code>'*'</code> 的正则表达式匹配。</p>
+
+<ul>
+	<li><code>'.'</code> 匹配任意单个字符</li>
+	<li><code>'*'</code> 匹配零个或多个前面的那一个元素</li>
+</ul>
+
+<p>所谓匹配，是要涵盖 <strong>整个 </strong>字符串 <code>s</code>的，而不是部分字符串。</p>
+ 
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "aa" p = "a"
+<strong>输出：</strong>false
+<strong>解释：</strong>"a" 无法匹配 "aa" 整个字符串。
+</pre>
+
+<p><strong>示例 2:</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "aa" p = "a*"
+<strong>输出：</strong>true
+<strong>解释：</strong>因为 '*' 代表可以匹配零个或多个前面的那一个元素, 在这里前面的元素就是 'a'。因此，字符串 "aa" 可被视为 'a' 重复了一次。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "ab" p = ".*"
+<strong>输出：</strong>true
+<strong>解释：</strong>".*" 表示可匹配零个或多个（'*'）任意字符（'.'）。
+</pre>
+
+<p><strong>示例 4：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "aab" p = "c*a*b"
+<strong>输出：</strong>true
+<strong>解释：</strong>因为 '*' 表示零个或多个，这里 'c' 为 0 个, 'a' 被重复一次。因此可以匹配字符串 "aab"。
+</pre>
+
+<p><strong>示例 5：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "mississippi" p = "mis*is*p*."
+<strong>输出：</strong>false</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>0 <= s.length <= 20</code></li>
+	<li><code>0 <= p.length <= 30</code></li>
+	<li><code>s</code> 可能为空，且只包含从 <code>a-z</code> 的小写字母。</li>
+	<li><code>p</code> 可能为空，且只包含从 <code>a-z</code> 的小写字母，以及字符 <code>.</code> 和 <code>*</code>。</li>
+	<li>保证每次出现字符 <code>*</code> 时，前面都匹配到有效的字符</li>
+</ul>
+<div><div>Related Topics</div><div><li>字符串</li><li>动态规划</li><li>回溯算法</li></div></div>\n<div><li>👍 2074</li><li>👎 0</li></div>

src/main/java/leetcode/editor/cn/SingleNumberIi.java

@@ -40,27 +40,35 @@ import java.util.HashMap;
 import java.util.Map;
 
 //137:只出现一次的数字 II
-public class SingleNumberIi{
+public class SingleNumberIi {
     public static void main(String[] args) {
         //测试代码
         Solution solution = new SingleNumberIi().new Solution();
     }
+
     //力扣代码
     //leetcode submit region begin(Prohibit modification and deletion)
-class Solution {
+    class Solution {
-    public int singleNumber(int[] nums) {
+        public int singleNumber(int[] nums) {
-        Map<Integer,Integer> map = new HashMap<>();
+            Map<Integer, Integer> map = new HashMap<>();
-        for (int num :nums){
+            for (int num : nums) {
-            map.put(num,map.getOrDefault(num,0)+1);
+                map.put(num, map.getOrDefault(num, 0) + 1);
-        }
-        for (int num:map.keySet()){
-            if(map.get(num)==1){
-                return num;
             }
+            for (int num : map.keySet()) {
+                if (map.get(num) == 1) {
+                    return num;
+                }
+            }
+            return 0;
+//            // 官方
+//            int ones = 0, twos = 0;
+//            for(int num : nums){
+//                ones = ones ^ num & ~twos;
+//                twos = twos ^ num & ~ones;
+//            }
+//            return ones;
         }
-        return 0;
     }
-}
 //leetcode submit region end(Prohibit modification and deletion)
 
 }