Merge remote-tracking branch 'origin/master'
# Conflicts: # src/main/java/leetcode/editor/cn/all.json
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LICENSE
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126
src/main/java/leetcode/editor/cn/RegularExpressionMatching.java
Normal file
126
src/main/java/leetcode/editor/cn/RegularExpressionMatching.java
Normal file
@ -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)
|
||||
|
||||
}
|
||||
@ -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>
|
||||
@ -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) {
|
||||
Map<Integer,Integer> map = new HashMap<>();
|
||||
for (int num :nums){
|
||||
map.put(num,map.getOrDefault(num,0)+1);
|
||||
Map<Integer, Integer> map = new HashMap<>();
|
||||
for (int num : nums) {
|
||||
map.put(num, map.getOrDefault(num, 0) + 1);
|
||||
}
|
||||
for (int num:map.keySet()){
|
||||
if(map.get(num)==1){
|
||||
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;
|
||||
}
|
||||
}
|
||||
}
|
||||
//leetcode submit region end(Prohibit modification and deletion)
|
||||
|
||||
}
|
||||
File diff suppressed because one or more lines are too long
Loading…
Reference in New Issue
Block a user