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废话不多说开始今天的解题Day~

53. Maximum Subarray

Question

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

A subarray is a contiguous part of an array.

Example

Example1

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

Example2

Input: nums = [1]
Output: 1

Example3

Input: nums = [5,4,-1,7,8]
Output: 23

Constraints

• 1 <= nums.length <= 3 * 104
• -105 <= nums[i] <= 105

Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

解题

题目

首先先简单的翻译一下题目
给定一个整数的阵列，从中找出总和最大的一个连续的子阵列(至少要有一个元素)，并回传总和最大的值。

Think

这个题目最暴力的情况就是用两个for loop去找最大总和，这样的时间复杂度会是O(n^2)。目标就先放在写出O(n)的演算法吧~

作法大致上是这样

• 因为要找的是总和最大的，所以如果今天找的阵列是[-2,1,4,3]，实际来跑一次流程看看。
• 一开始maxSum会是第一个值-2，currentSum则是代表每次子阵列的总和。
• 在Python第七行的这个if判断，是判断前面加总的值是不是小於0的，是的话代表不管接下来的元素是正是负都不会让currentSum更大。所以就让currentSum从头计算。
• 最後每次加完新的元素，再将currentSum跟maxSum中比较大的值存回maxSum。

Code

Python

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        maxSum = nums[0]
        currentSum = 0
        
        for num in nums:
            if currentSum < 0:
                currentSum = 0
            
            currentSum += num     
            maxSum = max(maxSum, currentSum)

        return maxSum

C

int maxSubArray(int* nums, int numsSize){
    int maxSum = nums[0];
    int currentSum = 0;

    for (int i=0 ; i<numsSize ; i++){
        if (currentSum < 0)
            currentSum = 0;

        currentSum += nums[i]; 
        maxSum = (maxSum > currentSum? maxSum : currentSum);
    }
        
    return maxSum;
}

Result

• Python

  • 

• C

  • 

大家明天见