feat: add python and java solutions to leetcode problem: No.0739

See https://leetcode-cn.com/problems/daily-temperatures

Author: yanglbme

README.md

@@ -99,6 +99,7 @@
 1. [用队列实现栈](/solution/0200-0299/0225.Implement%20Stack%20using%20Queues/README.md)
 1. [逆波兰表达式求值](/solution/0100-0199/0150.Evaluate%20Reverse%20Polish%20Notation/README.md)
 1. [最近的请求次数](/solution/0900-0999/0933.Number%20of%20Recent%20Calls/README.md)
+1. [每日温度](/solution/0700-0799/0739.Daily%20Temperatures/README.md)
 
 ### 动态规划
 

README_EN.md

@@ -94,6 +94,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 1. [Implement Stack using Queues](/solution/0200-0299/0225.Implement%20Stack%20using%20Queues/README_EN.md)
 1. [Evaluate Reverse Polish Notation](/solution/0100-0199/0150.Evaluate%20Reverse%20Polish%20Notation/README_EN.md)
 1. [Number of Recent Calls](/solution/0900-0999/0933.Number%20of%20Recent%20Calls/README_EN.md)
+1. [Daily Temperatures](/solution/0700-0799/0739.Daily%20Temperatures/README_EN.md)
 
 ### Dynamic Programming
 

solution/0700-0799/0739.Daily Temperatures/README.md

@@ -15,22 +15,55 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+栈实现，栈存放 T 中元素的的下标 i，结果用数组 res 存储。
+
+遍历 T，遍历到 `T[i]` 时：
+
+- 若栈不为空，并且栈顶元素小于 `T[i]` 时，弹出栈顶元素 j，并且 `res[j]` 赋值为 `i - j`。
+- 然后将 i 压入栈中。
+
+最后返回结果数组 res 即可。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def dailyTemperatures(self, T: List[int]) -> List[int]:
+        n = len(T)
+        res = [0 for _ in range(n)]
+        s = []
+        for i in range(n):
+            while s and T[s[-1]] < T[i]:
+                j = s.pop()
+                res[j] = i - j
+            s.append(i)
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int[] dailyTemperatures(int[] T) {
+        int n = T.length;
+        int[] res = new int[n];
+        Deque<Integer> s = new ArrayDeque<>();
+        for (int i = 0; i < n; ++i) {
+            while (!s.isEmpty() && T[s.peek()] < T[i]) {
+                int j = s.pop();
+                res[j] = i - j;
+            }
+            s.push(i);
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0700-0799/0739.Daily Temperatures/README_EN.md

@@ -29,13 +29,37 @@ Each temperature will be an integer in the range <code>[30, 100]</code>.
 ### **Python3**
 
 ```python
-
+class Solution:
+    def dailyTemperatures(self, T: List[int]) -> List[int]:
+        n = len(T)
+        res = [0 for _ in range(n)]
+        s = []
+        for i in range(n):
+            while s and T[s[-1]] < T[i]:
+                j = s.pop()
+                res[j] = i - j
+            s.append(i)
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int[] dailyTemperatures(int[] T) {
+        int n = T.length;
+        int[] res = new int[n];
+        Deque<Integer> s = new ArrayDeque<>();
+        for (int i = 0; i < n; ++i) {
+            while (!s.isEmpty() && T[s.peek()] < T[i]) {
+                int j = s.pop();
+                res[j] = i - j;
+            }
+            s.push(i);
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0700-0799/0739.Daily Temperatures/Solution.java

@@ -2,14 +2,14 @@ class Solution {
     public int[] dailyTemperatures(int[] T) {
         int n = T.length;
         int[] res = new int[n];
-        ArrayDeque<Integer> stack = new ArrayDeque<>();
+        Deque<Integer> s = new ArrayDeque<>();
         for (int i = 0; i < n; ++i) {
-            while (!stack.isEmpty() && T[stack.peek()] < T[i]) {
-                res[stack.peek()] = i - stack.peek();
-                stack.pop();
+            while (!s.isEmpty() && T[s.peek()] < T[i]) {
+                int j = s.pop();
+                res[j] = i - j;
             }
-            stack.push(i);
+            s.push(i);
         }
         return res;
     }
-}
+}

solution/0700-0799/0739.Daily Temperatures/Solution.py

@@ -0,0 +1,11 @@
+class Solution:
+    def dailyTemperatures(self, T: List[int]) -> List[int]:
+        n = len(T)
+        res = [0 for _ in range(n)]
+        s = []
+        for i in range(n):
+            while s and T[s[-1]] < T[i]:
+                j = s.pop()
+                res[j] = i - j
+            s.append(i)
+        return res

solution/0900-0999/0933.Number of Recent Calls/README.md

@@ -38,12 +38,12 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-<!-- tabs:start -->
-
 在第 1、100、3001、3002 这四个时间点分别进行了 ping 请求， 在 3001 秒的时候， 它前面的 3000 秒指的是区间 `[1,3001]`， 所以一共是有 `1、100、3001` 三个请求， t = 3002 的前 3000 秒指的是区间 `[2,3002]`, 所以有 `100、3001、3002` 三次请求。
 
 可以用队列实现。每次将 t 进入队尾，同时从队头开始依次移除小于 `t-3000` 的元素。然后返回队列的大小 `q.size()` 即可。
 
+<!-- tabs:start -->
+
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->