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

---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0053.Maximum%20Subarray/README.md
tags:
    - 数组
    - 分治
    - 动态规划
---

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# [53. 最大子数组和](https://leetcode.cn/problems/maximum-subarray)

[English Version](/solution/0000-0099/0053.Maximum%20Subarray/README_EN.md)

## 题目描述

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<p>给你一个整数数组 <code>nums</code> ，请你找出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。</p>

<p><strong><span data-keyword="subarray-nonempty">子数组 </span></strong>是数组中的一个连续部分。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>nums = [-2,1,-3,4,-1,2,1,-5,4]
<strong>输出：</strong>6
<strong>解释：</strong>连续子数组&nbsp;[4,-1,2,1] 的和最大，为&nbsp;6 。
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>nums = [1]
<strong>输出：</strong>1
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>nums = [5,4,-1,7,8]
<strong>输出：</strong>23
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>

<p><strong>进阶：</strong>如果你已经实现复杂度为 <code>O(n)</code> 的解法，尝试使用更为精妙的 <strong>分治法</strong> 求解。</p>

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## 解法

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### 方法一：动态规划

我们定义 $f[i]$ 表示以元素 $\textit{nums}[i]$ 为结尾的连续子数组的最大和，初始时 $f[0] = \textit{nums}[0]$，那么最终我们要求的答案即为 $\max_{0 \leq i < n} f[i]$。

考虑 $f[i]$，其中 $i \geq 1$，它的状态转移方程为：

$$
f[i] = \max(f[i - 1] + \textit{nums}[i], \textit{nums}[i])
$$

也即：

$$
f[i] = \max(f[i - 1], 0) + \textit{nums}[i]
$$

由于 $f[i]$ 只与 $f[i - 1]$ 有关系，因此我们可以只用一个变量 $f$ 来维护对于当前 $f[i]$ 的值是多少，然后进行状态转移即可。答案为 $\max_{0 \leq i < n} f$。

时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。

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#### Python3

```python
class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        ans = f = nums[0]
        for x in nums[1:]:
            f = max(f, 0) + x
            ans = max(ans, f)
        return ans
```

#### Java

```java
class Solution {
    public int maxSubArray(int[] nums) {
        int ans = nums[0];
        for (int i = 1, f = nums[0]; i < nums.length; ++i) {
            f = Math.max(f, 0) + nums[i];
            ans = Math.max(ans, f);
        }
        return ans;
    }
}
```

#### C++

```cpp
class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int ans = nums[0], f = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            f = max(f, 0) + nums[i];
            ans = max(ans, f);
        }
        return ans;
    }
};
```

#### Go

```go
func maxSubArray(nums []int) int {
	ans, f := nums[0], nums[0]
	for _, x := range nums[1:] {
		f = max(f, 0) + x
		ans = max(ans, f)
	}
	return ans
}
```

#### TypeScript

```ts
function maxSubArray(nums: number[]): number {
    let [ans, f] = [nums[0], nums[0]];
    for (let i = 1; i < nums.length; ++i) {
        f = Math.max(f, 0) + nums[i];
        ans = Math.max(ans, f);
    }
    return ans;
}
```

#### Rust

```rust
impl Solution {
    pub fn max_sub_array(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut ans = nums[0];
        let mut f = nums[0];
        for i in 1..n {
            f = f.max(0) + nums[i];
            ans = ans.max(f);
        }
        ans
    }
}
```

#### JavaScript

```js
/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function (nums) {
    let [ans, f] = [nums[0], nums[0]];
    for (let i = 1; i < nums.length; ++i) {
        f = Math.max(f, 0) + nums[i];
        ans = Math.max(ans, f);
    }
    return ans;
};
```

#### C#

```cs
public class Solution {
    public int MaxSubArray(int[] nums) {
        int ans = nums[0], f = nums[0];
        for (int i = 1; i < nums.Length; ++i) {
            f = Math.Max(f, 0) + nums[i];
            ans = Math.Max(ans, f);
        }
        return ans;
    }
}
```

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### 方法二

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#### Python3

```python
class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        def crossMaxSub(nums, left, mid, right):
            lsum = rsum = 0
            lmx = rmx = -inf
            for i in range(mid, left - 1, -1):
                lsum += nums[i]
                lmx = max(lmx, lsum)
            for i in range(mid + 1, right + 1):
                rsum += nums[i]
                rmx = max(rmx, rsum)
            return lmx + rmx

        def maxSub(nums, left, right):
            if left == right:
                return nums[left]
            mid = (left + right) >> 1
            lsum = maxSub(nums, left, mid)
            rsum = maxSub(nums, mid + 1, right)
            csum = crossMaxSub(nums, left, mid, right)
            return max(lsum, rsum, csum)

        left, right = 0, len(nums) - 1
        return maxSub(nums, left, right)
```

#### Java

```java
class Solution {
    public int maxSubArray(int[] nums) {
        return maxSub(nums, 0, nums.length - 1);
    }

    private int maxSub(int[] nums, int left, int right) {
        if (left == right) {
            return nums[left];
        }
        int mid = (left + right) >>> 1;
        int lsum = maxSub(nums, left, mid);
        int rsum = maxSub(nums, mid + 1, right);
        return Math.max(Math.max(lsum, rsum), crossMaxSub(nums, left, mid, right));
    }

    private int crossMaxSub(int[] nums, int left, int mid, int right) {
        int lsum = 0, rsum = 0;
        int lmx = Integer.MIN_VALUE, rmx = Integer.MIN_VALUE;
        for (int i = mid; i >= left; --i) {
            lsum += nums[i];
            lmx = Math.max(lmx, lsum);
        }
        for (int i = mid + 1; i <= right; ++i) {
            rsum += nums[i];
            rmx = Math.max(rmx, rsum);
        }
        return lmx + rmx;
    }
}
```

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