solution.md

# 最大子序和

<p>给定一个整数数组 <code>nums</code> ，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。</p><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>nums = [-2,1,-3,4,-1,2,1,-5,4]<strong><br />输出：</strong>6<strong><br />解释：</strong>连续子数组 [4,-1,2,1] 的和最大，为 6 。</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>nums = [1]<strong><br />输出：</strong>1</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>nums = [0]<strong><br />输出：</strong>0</pre><p><strong>示例 4：</strong></p><pre><strong>输入：</strong>nums = [-1]<strong><br />输出：</strong>-1</pre><p><strong>示例 5：</strong></p><pre><strong>输入：</strong>nums = [-100000]<strong><br />输出：</strong>-100000</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>	<li><code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code></li></ul><p> </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>

## template

```python
class Solution(object):
	def maxSubArray(self, nums):
		maxEndingHere = maxSofFar = nums[0]
		for i in range(1, len(nums)):
			maxEndingHere = max(maxEndingHere + nums[i], nums[i])
			maxSofFar = max(maxEndingHere, maxSofFar)
		return maxSofFar
# %%
s = Solution()
print(s.maxSubArray(nums = [-2,1,-3,4,-1,2,1,-5,4]))
```

## 答案

```python

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## 选项

### A

```python

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### B

```python

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### C

```python

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