feat: add solutions to leetcode problem: No.1365

Author: yanglbme

solution/0200-0299/0251.Flatten 2D Vector/README.md

@@ -46,7 +46,27 @@ iterator.hasNext(); // 返回 false
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class Vector2D:
 
+    def __init__(self, vec: List[List[int]]):
+        self.flatten = []
+        for item in vec:
+            for e in item:
+                self.flatten.append(e)
+        self.cur = -1
+
+    def next(self) -> int:
+        self.cur += 1
+        return self.flatten[self.cur]
+
+    def hasNext(self) -> bool:
+        return self.cur < len(self.flatten) - 1
+
+
+# Your Vector2D object will be instantiated and called as such:
+# obj = Vector2D(vec)
+# param_1 = obj.next()
+# param_2 = obj.hasNext()
 ```
 
 ### **Java**

solution/0200-0299/0251.Flatten 2D Vector/README_EN.md

@@ -44,7 +44,27 @@ iterator.hasNext(); // return false
 ### **Python3**
 
 ```python
+class Vector2D:
 
+    def __init__(self, vec: List[List[int]]):
+        self.flatten = []
+        for item in vec:
+            for e in item:
+                self.flatten.append(e)
+        self.cur = -1
+
+    def next(self) -> int:
+        self.cur += 1
+        return self.flatten[self.cur]
+
+    def hasNext(self) -> bool:
+        return self.cur < len(self.flatten) - 1
+
+
+# Your Vector2D object will be instantiated and called as such:
+# obj = Vector2D(vec)
+# param_1 = obj.next()
+# param_2 = obj.hasNext()
 ```
 
 ### **Java**

solution/0200-0299/0251.Flatten 2D Vector/Solution.java

@@ -0,0 +1,21 @@
+class Vector2D:
+
+    def __init__(self, vec: List[List[int]]):
+        self.flatten = []
+        for item in vec:
+            for e in item:
+                self.flatten.append(e)
+        self.cur = -1
+
+    def next(self) -> int:
+        self.cur += 1
+        return self.flatten[self.cur]
+
+    def hasNext(self) -> bool:
+        return self.cur < len(self.flatten) - 1
+
+
+# Your Vector2D object will be instantiated and called as such:
+# obj = Vector2D(vec)
+# param_1 = obj.next()
+# param_2 = obj.hasNext()

solution/0200-0299/0251.Flatten 2D Vector/Solution.py

@@ -50,22 +50,49 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+计数排序，求前缀和。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
+        cnt = [0] * 101
+        for num in nums:
+            cnt[num] += 1
+        for i in range(1, 101):
+            cnt[i] += cnt[i - 1]
+        res = []
+        for num in nums:
+            res.append(0 if num == 0 else cnt[num - 1])
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int[] smallerNumbersThanCurrent(int[] nums) {
+        int[] cnt = new int[101];
+        for (int e : nums) {
+            ++cnt[e];
+        }
+        for (int i = 1; i < 101; ++i) {
+            cnt[i] += cnt[i - 1];
+        }
+        int[] res = new int[nums.length];
+        for (int i = 0; i < nums.length; ++i) {
+            res[i] = nums[i] == 0 ? 0 : cnt[nums[i] - 1];
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/1300-1399/1365.How Many Numbers Are Smaller Than the Current Number/README.md

@@ -68,13 +68,38 @@ For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
 ### **Python3**
 
 ```python
-
+class Solution:
+    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
+        cnt = [0] * 101
+        for num in nums:
+            cnt[num] += 1
+        for i in range(1, 101):
+            cnt[i] += cnt[i - 1]
+        res = []
+        for num in nums:
+            res.append(0 if num == 0 else cnt[num - 1])
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int[] smallerNumbersThanCurrent(int[] nums) {
+        int[] cnt = new int[101];
+        for (int e : nums) {
+            ++cnt[e];
+        }
+        for (int i = 1; i < 101; ++i) {
+            cnt[i] += cnt[i - 1];
+        }
+        int[] res = new int[nums.length];
+        for (int i = 0; i < nums.length; ++i) {
+            res[i] = nums[i] == 0 ? 0 : cnt[nums[i] - 1];
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/1300-1399/1365.How Many Numbers Are Smaller Than the Current Number/README_EN.md

@@ -0,0 +1,16 @@
+class Solution {
+    public int[] smallerNumbersThanCurrent(int[] nums) {
+        int[] cnt = new int[101];
+        for (int e : nums) {
+            ++cnt[e];
+        }
+        for (int i = 1; i < 101; ++i) {
+            cnt[i] += cnt[i - 1];
+        }
+        int[] res = new int[nums.length];
+        for (int i = 0; i < nums.length; ++i) {
+            res[i] = nums[i] == 0 ? 0 : cnt[nums[i] - 1];
+        }
+        return res;
+    }
+}

solution/1300-1399/1365.How Many Numbers Are Smaller Than the Current Number/Solution.java

@@ -0,0 +1,11 @@
+class Solution:
+    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
+        cnt = [0] * 101
+        for num in nums:
+            cnt[num] += 1
+        for i in range(1, 101):
+            cnt[i] += cnt[i - 1]
+        res = []
+        for num in nums:
+            res.append(0 if num == 0 else cnt[num - 1])
+        return res