Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

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

思路：1. 取數組的第一個元素當做當前最大值curSum和最終最大值maxSum。然后從list第一個元素開始。如果num是正數，則將curSum+num賦給curSum。然后將curSum賦給maxSum。如果num是負數，則取num和curSum+num之間的大值(因為curSum+num的值有可能小于num的)，在將curSum和maxSum之間的大值賦給maxSum。

1. 依次從第一個元素遍歷list，然后比較nums[i]和nums[i] + nums[i-1]的大小，到第i個值的時候，這時候nums[i]是當前最大，如果繼續遍歷下去，就是比較nums[i+1]+nums[i]和nums[i+1]的大小，如果nums[i+1]是正值，則毫無疑問nums[i+1]+nums[i]大，如果nums[i+1]小于0，則取nums[i+1]+nums[i]和nums[i+1]的大值，依次循環下去，nums里面就會保留每次循環的大值，最后在其中選取最大值。

#!/usr/bin/env python
# -*- coding: UTF-8 -*-
class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        curSum = maxSum = nums[0]
        for num in nums[1:]:
            curSum = max(num, curSum + num)
            print num, "curSum:", curSum
            maxSum = max(curSum, maxSum)
            print "maxSum:", maxSum

        return maxSum

    def maxSubArray2(self, nums):
        for i in xrange(1, len(nums)):
            nums[i] = max(nums[i], nums[i] + nums[i-1])
            print i, " == ", nums
        return max(nums)



if __name__ == '__main__':
    sol = Solution()
    s = [-2, 1, -3, 4, -1, 2, 1, -5, 4]

    # print sol.maxSubArray(s)
    print sol.maxSubArray2(s)