feat: add python and java solutions to leetcode problem: No.0053

Author: yanglbme

README.md

@@ -131,6 +131,7 @@
 
 ### 动态规划
 
+- [最大子序和](/solution/0000-0099/0053.Maximum%20Subarray/README.md)
 - [打家劫舍](/solution/0100-0199/0198.House%20Robber/README.md)
 - [打家劫舍 II](/solution/0200-0299/0213.House%20Robber%20II/README.md)
 - [最长上升子序列](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README.md)

README_EN.md

@@ -120,6 +120,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 
 ### Dynamic Programming
 
+- [Maximum Subarray](/solution/0000-0099/0053.Maximum%20Subarray/README_EN.md)
 - [House Robber](/solution/0100-0199/0198.House%20Robber/README_EN.md)
 - [House Robber II](/solution/0200-0299/0213.House%20Robber%20II/README_EN.md)
 - [Longest Increasing Subsequence](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README_EN.md)

solution/0000-0099/0053.Maximum Subarray/README.md

@@ -22,22 +22,42 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+设 dp[i] 表示 `[0..i]` 中，以 nums[i] 结尾的最大子数组和，状态转移方程 `dp[i] = nums[i] + max(dp[i - 1], 0)`。
+
+由于 `dp[i]` 只与子问题 `dp[i-1]` 有关，故可以用一个变量 f 来表示。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def maxSubArray(self, nums: List[int]) -> int:
+        n = len(nums)
+        res = f = nums[0]
+        for i in range(1, n):
+            f = nums[i] + max(f, 0)
+            res = max(res, f)
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int maxSubArray(int[] nums) {
+        int f = nums[0], res = nums[0];
+        for (int i = 1, n = nums.length; i < n; ++i) {
+            f = nums[i] + Math.max(f, 0);
+            res = Math.max(res, f);
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0000-0099/0053.Maximum Subarray/README_EN.md

@@ -29,13 +29,29 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def maxSubArray(self, nums: List[int]) -> int:
+        n = len(nums)
+        res = f = nums[0]
+        for i in range(1, n):
+            f = nums[i] + max(f, 0)
+            res = max(res, f)
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int maxSubArray(int[] nums) {
+        int f = nums[0], res = nums[0];
+        for (int i = 1, n = nums.length; i < n; ++i) {
+            f = nums[i] + Math.max(f, 0);
+            res = Math.max(res, f);
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0000-0099/0053.Maximum Subarray/Solution.java

@@ -1,18 +1,10 @@
 class Solution {
     public int maxSubArray(int[] nums) {
-        int n = nums.length;
-        if (n == 1) {
-            return nums[0];
+        int f = nums[0], res = nums[0];
+        for (int i = 1, n = nums.length; i < n; ++i) {
+            f = nums[i] + Math.max(f, 0);
+            res = Math.max(res, f);
         }
-        int[] res = new int[n];
-        res[0] = nums[0];
-        int max = res[0];
-        for (int i = 1; i < n; ++i) {
-            res[i] = Math.max(res[i - 1] + nums[i], nums[i]);
-            max = Math.max(res[i], max);
-        }
-        
-        return max;
-        
+        return res;
     }
 }

solution/0000-0099/0053.Maximum Subarray/Solution.py

@@ -1,16 +1,8 @@
 class Solution:
-    def maxSubArray(self, nums):
-        """
-        :type nums: List[int]
-        :rtype: int
-        """
-        n=len(nums)
-        if n == 1:
-            return nums[0]
-        res=[0]*n
-        res[0]=nums[0]
-        max0=nums[0]
-        for i in range(1,n):
-            res[i]=max(res[i-1]+nums[i],nums[i])
-            max0=max(res[i],max0)
-        return max0
+    def maxSubArray(self, nums: List[int]) -> int:
+        n = len(nums)
+        res = f = nums[0]
+        for i in range(1, n):
+            f = nums[i] + max(f, 0)
+            res = max(res, f)
+        return res