python easy solution with time O(n) space O(1)

  • 1

    start from left one can find we only need to check weather current sum is larger or smaller than zero. If smaller, we can re-start at right position.

    class Solution(object):
        def maxSubArray(self, nums):
            :type nums: List[int]
            :rtype: int
            if not nums: return nums
            maxsum = nums[0]
            cursum = 0
            for i in range(len(nums)):
                cursum += nums[i]
                maxsum = max(maxsum,cursum)
                if cursum < 0:
                    cursum = 0
            return maxsum

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