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廢話不多說開始今天的解題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.

解題

題目

首先先簡單的翻譯一下題目
給定一個整數的陣列，從中找出總和最大的一個連續的子陣列(至少要有一個元素)，並回傳總和最大的值。

Think

這個題目最暴力的情況就是用兩個for loop去找最大總和，這樣的時間複雜度會是O(n^2)。目標就先放在寫出O(n)的演算法吧~

作法大致上是這樣

• 因為要找的是總和最大的，所以如果今天找的陣列是[-2,1,4,3]，實際來跑一次流程看看。
• 一開始maxSum會是第一個值-2，currentSum則是代表每次子陣列的總和。
• 在Python第七行的這個if判斷，是判斷前面加總的值是不是小於0的，是的話代表不管接下來的元素是正是負都不會讓currentSum更大。所以就讓currentSum從頭計算。
• 最後每次加完新的元素，再將currentSum跟maxSum中比較大的值存回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

  • 

大家明天見