feat: update leetcode solutions: No.0053. Maximum Subarray

Author: yanglbme

README.md

@@ -152,6 +152,7 @@
 
 ### 动态规划
 
+- [爬楼梯](/solution/0000-0099/0070.Climbing%20Stairs/README.md)
 - [接雨水](/solution/0000-0099/0042.Trapping%20Rain%20Water/README.md)
 - [最大子序和](/solution/0000-0099/0053.Maximum%20Subarray/README.md)
 - [乘积最大子序列](/solution/0100-0199/0152.Maximum%20Product%20Subarray/README.md)

README_EN.md

@@ -146,6 +146,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 
 ### Dynamic Programming
 
+- [Climbing Stairs](/solution/0000-0099/0070.Climbing%20Stairs/README_EN.md)
 - [Trapping Rain Water](/solution/0000-0099/0042.Trapping%20Rain%20Water/README_EN.md)
 - [Maximum Subarray](/solution/0000-0099/0053.Maximum%20Subarray/README_EN.md)
 - [Maximum Product Subarray](/solution/0100-0199/0152.Maximum%20Product%20Subarray/README_EN.md)

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

@@ -18,11 +18,13 @@
 
 <p>如果你已经实现复杂度为 O(<em>n</em>) 的解法，尝试使用更为精妙的分治法求解。</p>
 
+同[剑指 Offer 42. 连续子数组的最大和](/lcof/面试题42.%20连续子数组的最大和/README.md)。
+
 ## 解法
 
 <!-- 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)`。
+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。
 
@@ -60,6 +62,59 @@ class Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maxSubArray(vector<int>& nums) {
+        int f = nums[0], res = nums[0];
+        for (int i = 1; i < nums.size(); ++i) {
+            f = nums[i] + max(f, 0);
+            res = max(res, f);
+        }
+        return res;
+    }
+};
+```
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxSubArray = function (nums) {
+  let f = nums[0],
+    res = nums[0];
+  for (let i = 1; i < nums.length; ++i) {
+    f = nums[i] + Math.max(f, 0);
+    res = Math.max(res, f);
+  }
+  return res;
+};
+```
+
+### **Go**
+
+```go
+func maxSubArray(nums []int) int {
+    f, res := nums[0], nums[0]
+    for i := 1; i < len(nums); i++ {
+        if f > 0 {
+            f += nums[i]
+        } else {
+            f = nums[i]
+        }
+        if f > res {
+            res = f
+        }
+    }
+    return res
+}
+```
+
 ### **...**
 
 ```

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

@@ -54,6 +54,59 @@ class Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maxSubArray(vector<int>& nums) {
+        int f = nums[0], res = nums[0];
+        for (int i = 1; i < nums.size(); ++i) {
+            f = nums[i] + max(f, 0);
+            res = max(res, f);
+        }
+        return res;
+    }
+};
+```
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxSubArray = function (nums) {
+  let f = nums[0],
+    res = nums[0];
+  for (let i = 1; i < nums.length; ++i) {
+    f = nums[i] + Math.max(f, 0);
+    res = Math.max(res, f);
+  }
+  return res;
+};
+```
+
+### **Go**
+
+```go
+func maxSubArray(nums []int) int {
+    f, res := nums[0], nums[0]
+    for i := 1; i < len(nums); i++ {
+        if f > 0 {
+            f += nums[i]
+        } else {
+            f = nums[i]
+        }
+        if f > res {
+            res = f
+        }
+    }
+    return res
+}
+```
+
 ### **...**
 
 ```

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

@@ -1,74 +1,11 @@
-/*
- * @lc app=leetcode.cn id=53 lang=cpp
- *
- * [53] 最大子序和
- *
- * https://leetcode-cn.com/problems/maximum-subarray/description/
- *
- * algorithms
- * Easy (46.59%)
- * Likes:    2144
- * Dislikes: 0
- * Total Accepted:    270.6K
- * Total Submissions: 524.4K
- * Testcase Example:  '[-2,1,-3,4,-1,2,1,-5,4]'
- *
- * 给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。
- * 
- * 示例:
- * 
- * 输入: [-2,1,-3,4,-1,2,1,-5,4],
- * 输出: 6
- * 解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。
- * 
- * 
- * 进阶:
- * 
- * 如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的分治法求解。
- * 
- */
-
-#include <vector>
-
-using std::vector;
-
-// 贪心
-class Solution0 {
-   public:
-    int maxSubArray(vector<int>& nums) {
-        if (nums.size() == 0) {
-            return 0;
-        }
-        int result = nums[0];
-        int temp = nums[0];
-        for (int i = 1; i < nums.size(); i++) {
-            if (temp >= 0) {
-                temp += nums[i];
-            } else {
-                temp = nums[i];
-            }
-            result = std::max(result, temp);
-        }
-        return result;
-    }
-};
-
-// @lc code=start
-// 动态规划
 class Solution {
-   public:
+public:
     int maxSubArray(vector<int>& nums) {
-        if (nums.size() == 0) {
-            return 0;
-        }
-        vector<int> dp = vector<int>(nums.size());
-        dp[0] = nums[0];
-        int result = nums[0];
-        for (int i = 1; i < nums.size(); i++) {
-            dp[i] = std::max(dp[i - 1] + nums[i], nums[i]);
-            result = std::max(dp[i], result);
+        int f = nums[0], res = nums[0];
+        for (int i = 1; i < nums.size(); ++i) {
+            f = nums[i] + max(f, 0);
+            res = max(res, f);
         }
-        return result;
+        return res;
     }
-};
-// @lc code=end
+};

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

@@ -1,15 +1,14 @@
 func maxSubArray(nums []int) int {
-	ans := nums[0]
-	sum := 0
-	for _, n := range nums {
-		if sum > 0 {
-			sum += n
-		} else {
-			sum = n
-		}
-		if sum > ans {
-			ans = sum
-		}
-	}
-	return ans
-}
+    f, res := nums[0], nums[0]
+    for i := 1; i < len(nums); i++ {
+        if f > 0 {
+            f += nums[i]
+        } else {
+            f = nums[i]
+        }
+        if f > res {
+            res = f
+        }
+    }
+    return res
+}

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

@@ -1,10 +1,13 @@
-const maxSubArray = function (nums) {
-  if (nums.length === 0) return 0;
-  let ans = nums[0],
-    tmp = nums[0];
-  for (let i = 1; i < nums.length; i++) {
-    tmp = Math.max(tmp + nums[i], nums[i]);
-    ans = Math.max(ans, tmp);
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxSubArray = function (nums) {
+  let f = nums[0],
+    res = nums[0];
+  for (let i = 1; i < nums.length; ++i) {
+    f = nums[i] + Math.max(f, 0);
+    res = Math.max(res, f);
   }
-  return ans;
+  return res;
 };