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[Leetcode][Python]53: Maximum Subarray

时间:2015-03-30 22:30:08      阅读:195      评论:0      收藏:0      [点我收藏+]

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# -*- coding: utf8 -*-
‘‘‘
__author__ = ‘dabay.wang@gmail.com‘

53: Maximum Subarray
https://leetcode.com/problems/maximum-subarray/

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.

=== Comments by Dabay===
一维动态规划。
二分法的基本思想是:从中间分开,两边分别递归,同时处理跨界的情况。
http://www.cnblogs.com/springfor/p/3877058.html
‘‘‘

class Solution:
# @param A, a list of integers
# @return an integer
def maxSubArray(self, A):
if len(A) == 0:
return 0
max_so_far = max_ending_here = A[0]
for i in xrange(1, len(A)):
max_ending_here = max(max_ending_here + A[i], A[i])
max_so_far = max(max_so_far, max_ending_here)
return max_so_far


def main():
sol = Solution()
nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
print sol.maxSubArray(nums)


if __name__ == "__main__":
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)

[Leetcode][Python]53: Maximum Subarray

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原文地址:http://www.cnblogs.com/Dabay/p/4379079.html

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