feat: add solutions to lcof problem: No.42

Author: yanglbme

lcof/面试题42. 连续子数组的最大和/README.md

@@ -29,9 +29,29 @@
 
 ## 解法
 
-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)`。
+**方法一：动态规划**
 
-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。
+我们定义 $f[i]$ 表示以第 $i$ 个数结尾的「连续子数组的最大和」，那么很显然我们要求的答案就是：
+
+$$
+\max_{0 \leq i \leq n-1} f[i]
+$$
+
+那么我们如何求 $f[i]$ 呢？我们可以考虑 $nums[i]$ 单独成为一段还是加入 $f[i-1]$ 对应的那一段，这取决于 $nums[i]$ 和 $f[i-1] + nums[i]$ 哪个大，我们希望获得一个比较大的，于是可以写出这样的状态转移方程：
+
+$$
+f[i] = \max(f[i-1] + nums[i], nums[i])
+$$
+
+或者可以写成这样：
+
+$$
+f[i] = \max(f[i-1], 0) + nums[i]
+$$
+
+我们可以不用开一个数组来存储所有的计算结果，而是只用两个变量 $f$ 和 $ans$ 来维护对于每一个位置 $i$ 我们的最大值，这样我们可以省去空间复杂度的开销。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
 
 <!-- tabs:start -->
 
@@ -40,59 +60,42 @@
 ```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
+        ans, f = -inf, 0
+        for x in nums:
+            f = max(f, 0) + x
+            ans = max(ans, f)
+        return ans
 ```
 
 ### **Java**
 
 ```java
 class Solution {
     public int maxSubArray(int[] nums) {
-        int res = nums[0], f = 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 ans = Integer.MIN_VALUE;
+        int f = 0;
+        for (int x : nums) {
+            f = Math.max(f, 0) + x;
+            ans = Math.max(ans, f);
         }
-        return res;
+        return ans;
     }
 }
 ```
 
-### **JavaScript**
-
-```js
-/**
- * @param {number[]} nums
- * @return {number}
- */
-var maxSubArray = function (nums) {
-    let res = nums[0];
-    let f = nums[0];
-    for (let i = 1; i < nums.length; ++i) {
-        f = Math.max(f, 0) + nums[i];
-        res = Math.max(res, f);
-    }
-    return res;
-};
-```
-
 ### **C++**
 
 ```cpp
 class Solution {
 public:
     int maxSubArray(vector<int>& nums) {
-        int res = nums[0], f = nums[0];
-        for (int i = 1; i < nums.size(); ++i) {
-            f = max(f, 0) + nums[i];
-            res = max(res, f);
+        int ans = INT_MIN;
+        int f = 0;
+        for (int& x : nums) {
+            f = max(f, 0) + x;
+            ans = max(ans, f);
         }
-        return res;
+        return ans;
     }
 };
 ```
@@ -101,19 +104,38 @@ public:
 
 ```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
+	ans, f := -1000000000, 0
+	for _, x := range nums {
+		f = max(f, 0) + x
+		ans = max(ans, f)
+	}
+	return ans
 }
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxSubArray = function (nums) {
+    let ans = -1e10;
+    let f = 0;
+    for (const x of nums) {
+        f = Math.max(f, 0) + x;
+        ans = Math.max(ans, f);
+    }
+    return ans;
+};
 ```
 
 ### **TypeScript**
@@ -149,12 +171,13 @@ impl Solution {
 ```cs
 public class Solution {
     public int MaxSubArray(int[] nums) {
-        int pre = 0, maxAns = nums[0];
+        int ans = -1000000000;
+        int f = 0;
         foreach (int x in nums) {
-            pre = Math.Max(pre + x, x);
-            maxAns = Math.Max(maxAns, pre);
+            f = Math.Max(f, 0) + x;
+            ans = Math.Max(ans, f);
         }
-        return maxAns;
+        return ans;
     }
 }
 ```

lcof/面试题42. 连续子数组的最大和/Solution.cpp

@@ -1,11 +1,12 @@
 class Solution {
 public:
     int maxSubArray(vector<int>& nums) {
-        int res = nums[0], f = nums[0];
-        for (int i = 1; i < nums.size(); ++i) {
-            f = max(f, 0) + nums[i];
-            res = max(res, f);
+        int ans = INT_MIN;
+        int f = 0;
+        for (int& x : nums) {
+            f = max(f, 0) + x;
+            ans = max(ans, f);
         }
-        return res;
+        return ans;
     }
 };

lcof/面试题42. 连续子数组的最大和/Solution.cs

@@ -1,10 +1,11 @@
 public class Solution {
     public int MaxSubArray(int[] nums) {
-        int pre = 0, maxAns = nums[0];
+        int ans = -1000000000;
+        int f = 0;
         foreach (int x in nums) {
-            pre = Math.Max(pre + x, x);
-            maxAns = Math.Max(maxAns, pre);
+            f = Math.Max(f, 0) + x;
+            ans = Math.Max(ans, f);
         }
-        return maxAns;
+        return ans;
     }
 }

lcof/面试题42. 连续子数组的最大和/Solution.go

@@ -1,14 +1,15 @@
 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
+	ans, f := -1000000000, 0
+	for _, x := range nums {
+		f = max(f, 0) + x
+		ans = max(ans, f)
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
 }

lcof/面试题42. 连续子数组的最大和/Solution.java

@@ -1,10 +1,11 @@
 class Solution {
     public int maxSubArray(int[] nums) {
-        int res = nums[0], f = 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 ans = Integer.MIN_VALUE;
+        int f = 0;
+        for (int x : nums) {
+            f = Math.max(f, 0) + x;
+            ans = Math.max(ans, f);
         }
-        return res;
+        return ans;
     }
 }

lcof/面试题42. 连续子数组的最大和/Solution.js

@@ -3,11 +3,11 @@
  * @return {number}
  */
 var maxSubArray = function (nums) {
-    let res = nums[0];
-    let f = nums[0];
-    for (let i = 1; i < nums.length; ++i) {
-        f = Math.max(f, 0) + nums[i];
-        res = Math.max(res, f);
+    let ans = -1e10;
+    let f = 0;
+    for (const x of nums) {
+        f = Math.max(f, 0) + x;
+        ans = Math.max(ans, f);
     }
-    return res;
+    return ans;
 };

lcof/面试题42. 连续子数组的最大和/Solution.py

@@ -1,8 +1,7 @@
 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
+        ans, f = -inf, 0
+        for x in nums:
+            f = max(f, 0) + x
+            ans = max(ans, f)
+        return ans