題目：
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 1:
Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Example 2:
Input: nums = [1]
Output: 1
Example 3:
Input: nums = [5,4,-1,7,8]
Output: 23
Constraints:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
我自己用了暴力求解法竣付，通過兩層循環(huán)嵌套牙勘，遍歷數(shù)組中所有的子序并求和得到最大的子序和。這種方法時(shí)間復(fù)雜度大极谊。看最佳答案看到一條：
nums = [-1, 1, -2,4, -3, 8]
maxSub = nums[0]
curSum = 0
for n in nums:
if curSum < 0:
curSum = 0
curSum += n
print (curSum)
maxSub = max(maxSub, curSum)
print (maxSub)
一開始想了半天琅翻，沒想通為什么curSum<0時(shí)可以直接賦值為0.后來突然想到發(fā)現(xiàn)其實(shí)很簡單隐锭，當(dāng)前面的子序都小于0，那么也沒必要在往后找的時(shí)候加上前面的子序论悴，只會(huì)越加越小掖棉，所以可以直接等于0.
