[pull] main from doocs:main

solution/0000-0099/0013.Roman to Integer/README.md

@@ -186,6 +186,28 @@ func romanToInt(s string) int {
 }
 ```
 
+### **PHP**
+
+```php
+class Solution {
+    /**
+     * @param String $s
+     * @return Integer
+     */
+    function romanToInt($s) {
+        $hashmap = array('I' => 1, 'V' => 5, 'X' => 10, 'L' => 50, 'C' => 100, 'D' => 500, 'M' => 1000);
+        $rs = 0;
+        for ($i = 0; $i < strlen($s); $i++) {
+            $left = $hashmap[$s[$i]];
+            $right = $hashmap[$s[$i + 1]];
+            if ($left >= $right) $rs += $left;
+            else $rs -= $left;
+        }
+        return $rs;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0000-0099/0013.Roman to Integer/README_EN.md

@@ -164,6 +164,28 @@ func romanToInt(s string) int {
 }
 ```
 
+### **PHP**
+
+```php
+class Solution {
+    /**
+     * @param String $s
+     * @return Integer
+     */
+    function romanToInt($s) {
+        $hashmap = array('I' => 1, 'V' => 5, 'X' => 10, 'L' => 50, 'C' => 100, 'D' => 500, 'M' => 1000);
+        $rs = 0;
+        for ($i = 0; $i < strlen($s); $i++) {
+            $left = $hashmap[$s[$i]];
+            $right = $hashmap[$s[$i + 1]];
+            if ($left >= $right) $rs += $left;
+            else $rs -= $left;
+        }
+        return $rs;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0000-0099/0013.Roman to Integer/Solution.php

@@ -0,0 +1,17 @@
+class Solution {
+    /**
+     * @param String $s
+     * @return Integer
+     */
+    function romanToInt($s) {
+        $hashmap = array('I' => 1, 'V' => 5, 'X' => 10, 'L' => 50, 'C' => 100, 'D' => 500, 'M' => 1000);
+        $rs = 0;
+        for ($i = 0; $i < strlen($s); $i++) {
+            $left = $hashmap[$s[$i]];
+            $right = $hashmap[$s[$i + 1]];
+            if ($left >= $right) $rs += $left;
+            else $rs -= $left;
+        }
+        return $rs;
+    }
+}

solution/0100-0199/0125.Valid Palindrome/README.md

@@ -310,7 +310,7 @@ func verify(ch byte) bool {
 
 ### **PHP**
 
-```php 
+```php
 class Solution {
     /**
      * @param String $s

solution/0100-0199/0125.Valid Palindrome/README_EN.md

@@ -298,7 +298,7 @@ func verify(ch byte) bool {
 
 ### **PHP**
 
-```php 
+```php
 class Solution {
     /**
      * @param String $s

solution/0100-0199/0188.Best Time to Buy and Sell Stock IV/README_EN.md

@@ -6,7 +6,7 @@
 
 <p>You are given an integer array <code>prices</code> where <code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day, and an integer <code>k</code>.</p>
 
-<p>Find the maximum profit you can achieve. You may complete at most <code>k</code> transactions.</p>
+<p>Find the maximum profit you can achieve. You may complete at most <code>k</code> transactions: i.e. you may buy at most <code>k</code> times and sell at most <code>k</code> times.</p>
 
 <p><strong>Note:</strong> You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).</p>
 

solution/0200-0299/0211.Design Add and Search Words Data Structure/README_EN.md

@@ -42,7 +42,7 @@ wordDictionary.search("b.."); // return True
 	<li><code>1 &lt;= word.length &lt;= 25</code></li>
 	<li><code>word</code> in <code>addWord</code> consists of lowercase English letters.</li>
 	<li><code>word</code> in <code>search</code> consist of <code>&#39;.&#39;</code> or lowercase English letters.</li>
-	<li>There will be at most <code>3</code> dots in <code>word</code> for <code>search</code> queries.</li>
+	<li>There will be at most <code>2</code> dots in <code>word</code> for <code>search</code> queries.</li>
 	<li>At most <code>10<sup>4</sup></code> calls will be made to <code>addWord</code> and <code>search</code>.</li>
 </ul>
 

solution/1700-1799/1734.Decode XORed Permutation/README.md

@@ -41,6 +41,12 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：位运算**
+
+我们注意到，数组 $perm$ 是前 $n$ 个正整数的排列，因此 $perm$ 的所有元素的异或和为 $1 \oplus 2 \oplus \cdots \oplus n$，记为 $a$。而 $encode[i]=perm[i] \oplus perm[i+1]$，如果我们将 $encode[0],encode[2],\cdots,encode[n-3]$ 的所有元素的异或和记为 $b$，则 $perm[n-1]=a \oplus b$。知道了 $perm$ 的最后一个元素，我们就可以通过逆序遍历数组 $encode$ 求出 $perm$ 的所有元素。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 $perm$ 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -54,12 +60,13 @@ class Solution:
         a = b = 0
         for i in range(0, n - 1, 2):
             a ^= encoded[i]
-        for i in range(n + 1):
+        for i in range(1, n + 1):
             b ^= i
-        ans = [a ^ b]
-        for e in encoded[::-1]:
-            ans.append(ans[-1] ^ e)
-        return ans[::-1]
+        perm = [0] * n
+        perm[-1] = a ^ b
+        for i in range(n - 2, -1, -1):
+            perm[i] = encoded[i] ^ perm[i + 1]
+        return perm
 ```
 
 ### **Java**
@@ -68,23 +75,21 @@ class Solution:
 
 ```java
 class Solution {
-
     public int[] decode(int[] encoded) {
         int n = encoded.length + 1;
-        int[] ans = new int[n];
-        int a = 0;
-        int b = 0;
+        int a = 0, b = 0;
         for (int i = 0; i < n - 1; i += 2) {
             a ^= encoded[i];
         }
-        for (int i = 0; i < n + 1; ++i) {
+        for (int i = 1; i <= n; ++i) {
             b ^= i;
         }
-        ans[n - 1] = a ^ b;
+        int[] perm = new int[n];
+        perm[n - 1] = a ^ b;
         for (int i = n - 2; i >= 0; --i) {
-            ans[i] = ans[i + 1] ^ encoded[i];
+            perm[i] = encoded[i] ^ perm[i + 1];
         }
-        return ans;
+        return perm;
     }
 }
 ```
@@ -96,13 +101,19 @@ class Solution {
 public:
     vector<int> decode(vector<int>& encoded) {
         int n = encoded.size() + 1;
-        vector<int> ans(n);
         int a = 0, b = 0;
-        for (int i = 0; i < n - 1; i += 2) a ^= encoded[i];
-        for (int i = 0; i < n + 1; ++i) b ^= i;
-        ans[n - 1] = a ^ b;
-        for (int i = n - 2; i >= 0; --i) ans[i] = ans[i + 1] ^ encoded[i];
-        return ans;
+        for (int i = 0; i < n - 1; i += 2) {
+            a ^= encoded[i];
+        }
+        for (int i = 1; i <= n; ++i) {
+            b ^= i;
+        }
+        vector<int> perm(n);
+        perm[n - 1] = a ^ b;
+        for (int i = n - 2; ~i; --i) {
+            perm[i] = encoded[i] ^ perm[i + 1];
+        }
+        return perm;
     }
 };
 ```
@@ -112,19 +123,19 @@ public:
 ```go
 func decode(encoded []int) []int {
 	n := len(encoded) + 1
-	ans := make([]int, n)
 	a, b := 0, 0
 	for i := 0; i < n-1; i += 2 {
 		a ^= encoded[i]
 	}
-	for i := 0; i < n+1; i++ {
+	for i := 1; i <= n; i++ {
 		b ^= i
 	}
-	ans[n-1] = a ^ b
+	perm := make([]int, n)
+	perm[n-1] = a ^ b
 	for i := n - 2; i >= 0; i-- {
-		ans[i] = ans[i+1] ^ encoded[i]
+		perm[i] = encoded[i] ^ perm[i+1]
 	}
-	return ans
+	return perm
 }
 ```
 

solution/1700-1799/1734.Decode XORed Permutation/README_EN.md

@@ -48,12 +48,13 @@ class Solution:
         a = b = 0
         for i in range(0, n - 1, 2):
             a ^= encoded[i]
-        for i in range(n + 1):
+        for i in range(1, n + 1):
             b ^= i
-        ans = [a ^ b]
-        for e in encoded[::-1]:
-            ans.append(ans[-1] ^ e)
-        return ans[::-1]
+        perm = [0] * n
+        perm[-1] = a ^ b
+        for i in range(n - 2, -1, -1):
+            perm[i] = encoded[i] ^ perm[i + 1]
+        return perm
 ```
 
 ### **Java**
@@ -62,20 +63,19 @@ class Solution:
 class Solution {
     public int[] decode(int[] encoded) {
         int n = encoded.length + 1;
-        int[] ans = new int[n];
-        int a = 0;
-        int b = 0;
+        int a = 0, b = 0;
         for (int i = 0; i < n - 1; i += 2) {
             a ^= encoded[i];
         }
-        for (int i = 0; i < n + 1; ++i) {
+        for (int i = 1; i <= n; ++i) {
             b ^= i;
         }
-        ans[n - 1] = a ^ b;
+        int[] perm = new int[n];
+        perm[n - 1] = a ^ b;
         for (int i = n - 2; i >= 0; --i) {
-            ans[i] = ans[i + 1] ^ encoded[i];
+            perm[i] = encoded[i] ^ perm[i + 1];
         }
-        return ans;
+        return perm;
     }
 }
 ```
@@ -87,13 +87,19 @@ class Solution {
 public:
     vector<int> decode(vector<int>& encoded) {
         int n = encoded.size() + 1;
-        vector<int> ans(n);
         int a = 0, b = 0;
-        for (int i = 0; i < n - 1; i += 2) a ^= encoded[i];
-        for (int i = 0; i < n + 1; ++i) b ^= i;
-        ans[n - 1] = a ^ b;
-        for (int i = n - 2; i >= 0; --i) ans[i] = ans[i + 1] ^ encoded[i];
-        return ans;
+        for (int i = 0; i < n - 1; i += 2) {
+            a ^= encoded[i];
+        }
+        for (int i = 1; i <= n; ++i) {
+            b ^= i;
+        }
+        vector<int> perm(n);
+        perm[n - 1] = a ^ b;
+        for (int i = n - 2; ~i; --i) {
+            perm[i] = encoded[i] ^ perm[i + 1];
+        }
+        return perm;
     }
 };
 ```
@@ -103,19 +109,19 @@ public:
 ```go
 func decode(encoded []int) []int {
 	n := len(encoded) + 1
-	ans := make([]int, n)
 	a, b := 0, 0
 	for i := 0; i < n-1; i += 2 {
 		a ^= encoded[i]
 	}
-	for i := 0; i < n+1; i++ {
+	for i := 1; i <= n; i++ {
 		b ^= i
 	}
-	ans[n-1] = a ^ b
+	perm := make([]int, n)
+	perm[n-1] = a ^ b
 	for i := n - 2; i >= 0; i-- {
-		ans[i] = ans[i+1] ^ encoded[i]
+		perm[i] = encoded[i] ^ perm[i+1]
 	}
-	return ans
+	return perm
 }
 ```
 

solution/1700-1799/1734.Decode XORed Permutation/Solution.cpp

@@ -2,12 +2,18 @@ class Solution {
 public:
     vector<int> decode(vector<int>& encoded) {
         int n = encoded.size() + 1;
-        vector<int> ans(n);
         int a = 0, b = 0;
-        for (int i = 0; i < n - 1; i += 2) a ^= encoded[i];
-        for (int i = 0; i < n + 1; ++i) b ^= i;
-        ans[n - 1] = a ^ b;
-        for (int i = n - 2; i >= 0; --i) ans[i] = ans[i + 1] ^ encoded[i];
-        return ans;
+        for (int i = 0; i < n - 1; i += 2) {
+            a ^= encoded[i];
+        }
+        for (int i = 1; i <= n; ++i) {
+            b ^= i;
+        }
+        vector<int> perm(n);
+        perm[n - 1] = a ^ b;
+        for (int i = n - 2; ~i; --i) {
+            perm[i] = encoded[i] ^ perm[i + 1];
+        }
+        return perm;
     }
 };

solution/1700-1799/1734.Decode XORed Permutation/Solution.go

@@ -1,16 +1,16 @@
 func decode(encoded []int) []int {
 	n := len(encoded) + 1
-	ans := make([]int, n)
 	a, b := 0, 0
 	for i := 0; i < n-1; i += 2 {
 		a ^= encoded[i]
 	}
-	for i := 0; i < n+1; i++ {
+	for i := 1; i <= n; i++ {
 		b ^= i
 	}
-	ans[n-1] = a ^ b
+	perm := make([]int, n)
+	perm[n-1] = a ^ b
 	for i := n - 2; i >= 0; i-- {
-		ans[i] = ans[i+1] ^ encoded[i]
+		perm[i] = encoded[i] ^ perm[i+1]
 	}
-	return ans
+	return perm
 }

solution/1700-1799/1734.Decode XORed Permutation/Solution.java

@@ -1,19 +1,18 @@
 class Solution {
     public int[] decode(int[] encoded) {
         int n = encoded.length + 1;
-        int[] ans = new int[n];
-        int a = 0;
-        int b = 0;
+        int a = 0, b = 0;
         for (int i = 0; i < n - 1; i += 2) {
             a ^= encoded[i];
         }
-        for (int i = 0; i < n + 1; ++i) {
+        for (int i = 1; i <= n; ++i) {
             b ^= i;
         }
-        ans[n - 1] = a ^ b;
+        int[] perm = new int[n];
+        perm[n - 1] = a ^ b;
         for (int i = n - 2; i >= 0; --i) {
-            ans[i] = ans[i + 1] ^ encoded[i];
+            perm[i] = encoded[i] ^ perm[i + 1];
         }
-        return ans;
+        return perm;
     }
 }