feat: add solutions to lc problems: No.1859,1860,1861

Author: yanglbme

solution/1800-1899/1859.Sorting the Sentence/README.md

@@ -45,7 +45,6 @@
 	<li><code>s</code> 不包含任何前导或者后缀空格。</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
@@ -57,15 +56,32 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def sortSentence(self, s: str) -> str:
+        words = s.split(' ')
+        arr = [None] * len(words)
+        for word in words:
+            idx = int(word[-1]) - 1
+            arr[idx] = word[:-1]
+        return ' '.join(arr)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public String sortSentence(String s) {
+        String[] words = s.split(" ");
+        String[] arr = new String[words.length];
+        for (String word : words) {
+            int idx = word.charAt(word.length() - 1) - '0' - 1;
+            arr[idx] = word.substring(0, word.length() - 1);
+        }
+        return String.join(" ", arr);
+    }
+}
 ```
 
 ### **...**

solution/1800-1899/1859.Sorting the Sentence/README_EN.md

@@ -6,28 +6,18 @@
 
 <p>A <strong>sentence</strong> is a list of words that are separated by a single space with no leading or trailing spaces. Each word consists of lowercase and uppercase English letters.</p>
 
-
-
 <p>A sentence can be <strong>shuffled</strong> by appending the <strong>1-indexed word position</strong> to each word then rearranging the words in the sentence.</p>
 
-
-
 <ul>
 	<li>For example, the sentence <code>&quot;This is a sentence&quot;</code> can be shuffled as <code>&quot;sentence4 a3 is2 This1&quot;</code> or <code>&quot;is2 sentence4 This1 a3&quot;</code>.</li>
 </ul>
 
-
-
 <p>Given a <strong>shuffled sentence</strong> <code>s</code> containing no more than <code>9</code> words, reconstruct and return <em>the original sentence</em>.</p>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Example 1:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> s = &quot;is2 sentence4 This1 a3&quot;
@@ -38,12 +28,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 2:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> s = &quot;Myself2 Me1 I4 and3&quot;
@@ -54,14 +40,10 @@
 
 </pre>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Constraints:</strong></p>
 
-
-
 <ul>
 	<li><code>2 &lt;= s.length &lt;= 200</code></li>
 	<li><code>s</code> consists of lowercase and uppercase English letters, spaces, and digits from <code>1</code> to <code>9</code>.</li>
@@ -77,13 +59,30 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def sortSentence(self, s: str) -> str:
+        words = s.split(' ')
+        arr = [None] * len(words)
+        for word in words:
+            idx = int(word[-1]) - 1
+            arr[idx] = word[:-1]
+        return ' '.join(arr)
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public String sortSentence(String s) {
+        String[] words = s.split(" ");
+        String[] arr = new String[words.length];
+        for (String word : words) {
+            int idx = word.charAt(word.length() - 1) - '0' - 1;
+            arr[idx] = word.substring(0, word.length() - 1);
+        }
+        return String.join(" ", arr);
+    }
+}
 ```
 
 ### **...**

solution/1800-1899/1859.Sorting the Sentence/Solution.java

@@ -0,0 +1,11 @@
+class Solution {
+    public String sortSentence(String s) {
+        String[] words = s.split(" ");
+        String[] arr = new String[words.length];
+        for (String word : words) {
+            int idx = word.charAt(word.length() - 1) - '0' - 1;
+            arr[idx] = word.substring(0, word.length() - 1);
+        }
+        return String.join(" ", arr);
+    }
+}

solution/1800-1899/1859.Sorting the Sentence/Solution.py

@@ -0,0 +1,8 @@
+class Solution:
+    def sortSentence(self, s: str) -> str:
+        words = s.split(' ')
+        arr = [None] * len(words)
+        for word in words:
+            idx = int(word[-1]) - 1
+            arr[idx] = word[:-1]
+        return ' '.join(arr)

solution/1800-1899/1860.Incremental Memory Leak/README.md

@@ -45,7 +45,6 @@
 	<li><code>0 &lt;= memory1, memory2 &lt;= 2<sup>31</sup> - 1</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
@@ -57,15 +56,47 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def memLeak(self, memory1: int, memory2: int) -> List[int]:
+        i = 1
+        while 1:
+            if memory1 >= memory2:
+                if memory1 < i:
+                    break
+                memory1 -= i
+            else:
+                if memory2 < i:
+                    break
+                memory2 -= i
+            i += 1
+        return [i, memory1, memory2]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int[] memLeak(int memory1, int memory2) {
+        int i = 1;
+        while (true) {
+            if (memory1 >= memory2) {
+                if (memory1 < i) {
+                    break;
+                }
+                memory1 -= i;
+            } else {
+                if (memory2 < i) {
+                    break;
+                }
+                memory2 -= i;
+            }
+            ++i;
+        }
+        return new int[]{i, memory1, memory2};
+    }
+}
 ```
 
 ### **...**

solution/1800-1899/1860.Incremental Memory Leak/README_EN.md

@@ -43,21 +43,52 @@
 	<li><code>0 &lt;= memory1, memory2 &lt;= 2<sup>31</sup> - 1</code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def memLeak(self, memory1: int, memory2: int) -> List[int]:
+        i = 1
+        while 1:
+            if memory1 >= memory2:
+                if memory1 < i:
+                    break
+                memory1 -= i
+            else:
+                if memory2 < i:
+                    break
+                memory2 -= i
+            i += 1
+        return [i, memory1, memory2]
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int[] memLeak(int memory1, int memory2) {
+        int i = 1;
+        while (true) {
+            if (memory1 >= memory2) {
+                if (memory1 < i) {
+                    break;
+                }
+                memory1 -= i;
+            } else {
+                if (memory2 < i) {
+                    break;
+                }
+                memory2 -= i;
+            }
+            ++i;
+        }
+        return new int[]{i, memory1, memory2};
+    }
+}
 ```
 
 ### **...**

solution/1800-1899/1860.Incremental Memory Leak/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public int[] memLeak(int memory1, int memory2) {
+        int i = 1;
+        while (true) {
+            if (memory1 >= memory2) {
+                if (memory1 < i) {
+                    break;
+                }
+                memory1 -= i;
+            } else {
+                if (memory2 < i) {
+                    break;
+                }
+                memory2 -= i;
+            }
+            ++i;
+        }
+        return new int[]{i, memory1, memory2};
+    }
+}

solution/1800-1899/1860.Incremental Memory Leak/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def memLeak(self, memory1: int, memory2: int) -> List[int]:
+        i = 1
+        while 1:
+            if memory1 >= memory2:
+                if memory1 < i:
+                    break
+                memory1 -= i
+            else:
+                if memory2 < i:
+                    break
+                memory2 -= i
+            i += 1
+        return [i, memory1, memory2]

solution/1800-1899/1861.Rotating the Box/README.md

@@ -69,27 +69,79 @@
 	<li><code>box[i][j]</code> 只可能是 <code>'#'</code> ，<code>'*'</code> 或者 <code>'.'</code> 。</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+先旋转，再挪动箱子。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def rotateTheBox(self, box: List[List[str]]) -> List[List[str]]:
+        m, n = len(box), len(box[0])
+        res = [[None] * m for _ in range(n)]
+        for i in range(m):
+            for j in range(n):
+                res[j][m - i - 1] = box[i][j]
+        for j in range(m):
+            q = collections.deque()
+            for i in range(n - 1, -1, -1):
+                if res[i][j] == '*':
+                    q.clear()
+                    continue
+                if res[i][j] == '.':
+                    q.append(i)
+                else:
+                    if not q:
+                        continue
+                    res[q.popleft()][j] = '#'
+                    res[i][j] = '.'
+                    q.append(i)
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public char[][] rotateTheBox(char[][] box) {
+        int m = box.length, n = box[0].length;
+        char[][] res = new char[n][m];
+        for (int i = 0; i < m; ++i) {
+            for (int j = 0; j < n; ++j) {
+                res[j][m - i - 1] = box[i][j];
+            }
+        }
+        for (int j = 0; j < m; ++j) {
+            Deque<Integer> q = new ArrayDeque<>();
+            for (int i = n - 1; i >= 0; --i) {
+                if (res[i][j] == '*') {
+                    q.clear();
+                    continue;
+                }
+                if (res[i][j] == '.') {
+                    q.offer(i);
+                } else {
+                    if (q.isEmpty()) {
+                        continue;
+                    }
+                    res[q.poll()][j] = '#';
+                    res[i][j] = '.';
+                    q.offer(i);
+                }
+            }
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**