GoCalf Calf 的自留地 · 昙花一现

53. Maximum Subarray

Calf

Problem

Given an integer array nums, find the subarray with the largest sum, and return its sum.

A subarray is a contiguous non-empty sequence of elements within an array.

https://leetcode.com/problems/maximum-subarray/

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: The subarray [4,-1,2,1] has the largest sum 6.

Example 2:

Input: nums = [1]
Output: 1
Explanation: The subarray [1] has the largest sum 1.

Example 3:

Input: nums = [5,4,-1,7,8]
Output: 23
Explanation: The subarray [5,4,-1,7,8] has the largest sum 23.

Constraints:

  • 1 <= nums.length <= 10⁵
  • -10⁴ <= nums[i] <= 10⁴

Test Cases

python 
class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
solution_test.py下载完整源码
py 
import pytest

from solution import Solution
from solution2 import Solution as Solution2
from solution3 import Solution as Solution3


@pytest.mark.parametrize('nums, expected', [
    ([-2,1,-3,4,-1,2,1,-5,4], 6),
    ([1], 1),
    ([5,4,-1,7,8], 23),
])
class Test:
    def test_solution(self, nums, expected):
        sol = Solution()
        assert sol.maxSubArray(nums) == expected

    def test_solution2(self, nums, expected):
        sol = Solution2()
        assert sol.maxSubArray(nums) == expected

    def test_solution3(self, nums, expected):
        sol = Solution3()
        assert sol.maxSubArray(nums) == expected

Thoughts

如果一个 subarray 的和大于 0，把它与下一个数相加的结果一定比下一个数自身大。反之，应该从下一个数开启新的 subarray。

记录当前 subarray 的总和，如果把下一个数加入进来，总和仍然大于 0，就可以继续。否则把 subarray 清空，总和恢复为 0，从再下一个数开始尝试建立新的 subarray。过程中记录见到的最大的总和。

Code

solution.py下载完整源码
py 
from typing import List


class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        largest = nums[0]
        s = 0
        for v in nums:
            s += v
            largest = max(largest, s)
            s = max(s, 0)

        return largest

Another DP

在 1186. Maximum Subarray Sum with One Deletion 中提到上边计算逻辑的状态定义和状态转移：s(i) 表示以 i 为右端点的最大 subarray 和（至少包含 nums[i]），即：

{s(0)=nums[0]s(i)=max⁡{s(i−1),0}+nums[i] \begin{cases} s(0)=nums[0] \\ s(i)=\max\{s(i-1),0\}+nums[i] \end{cases}

那么 nums 的最大 subarray 和为 largest=max⁡0≤i<ns(i)largest=\max_{0\le i<n}{s(i)}。

上边代码在关于 i 的循环开始的时候计算了 s(i)，在循环结束前把 max{s(i), 0} 先算好，到 i + 1 的循环里直接用。也可以写成如下形式，逻辑是完全一致的，跟公式更接近一些：

solution2.py下载完整源码
py 
class Solution:
    def maxSubArray(self, nums: list[int]) -> int:
        max2 = lambda a, b: a if a >= b else b
        s = largest = nums[0]
        for i in range(1, len(nums)):
            s = max2(s, 0) + nums[i]
            largest = max2(largest, s)

        return largest

为了能理解 3410. Maximize Subarray Sum After Removing All Occurrences of One Element，这里对上述 DP 做一些调整，可以得到相同的结果。

这次直接记录数组的前缀和，即 ps(i) = Σnums[0...i]，同时用 low(i) 记录 ps(0)...ps(i) 的（小于等于零的）最小值，即：

ps(i)=∑j=0inums[j]low(i)=min⁡{0,min⁡0≤j≤ips(j)} \begin{array}{rcl} ps(i) & = &\sum_{j=0}^{i}nums[j] \\ low(i) & = & \min\{0, \min_{0\le j\le i}ps(j)\} \end{array}

显然 s(i)=ps(i)−low(i−1)s(i)=ps(i)-low(i-1)。可得 largest=max⁡0≤i<ns(i)=max⁡0≤i<n{ps(i)−low(i−1)}largest=\max_{0\le i<n}{s(i)}=\max_{0\le i<n}\{ps(i)-low(i-1)\}（令 low(-1) = 0）。

solution3.py下载完整源码
py 
class Solution:
    def maxSubArray(self, nums: list[int]) -> int:
        max2 = lambda a, b: a if a >= b else b
        min2 = lambda a, b: a if a <= b else b
        ps = largest = nums[0]
        low = 0
        for i in range(1, len(nums)):
            low = min2(low, ps) # low(i-1) = min(low(i-2), ps(i-1))
            ps += nums[i] # ps(i) = ps(i-1) + nums[i]
            largest = max2(largest, ps - low) # s(i) = ps(i) - low(i-1)

        return largest

这样改造之后，problem 3410 比较容易在此基础上进行扩展。

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