LeetCode Notes_#53 Maximum Subarray

LeetCode Notes_#53 Maximum Subarray

LeetCode 

Contents

• 题目
• 思路和解答
  • 思路
  • 解答
    • 需要注意的点

题目

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

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.

思路和解答

思路

第一想法就是最简单的，直接暴力遍历整个数组，维护一个max变量，然后不断更新max的值，尝试一下

nums=[1,2,3,4,5]
nums[1:]

[2, 3, 4, 5]

#暴力算法，比较长的用例会超时....
maxSum=[-2,1,-3,4,-1,2,1,-5,4]
maxSum=nums[0]
tmpSum=0
for i in range(len(nums)):
    for j in nums[i:]:
        tmpSum=tmpSum+j
        maxSum=max(maxSum,tmpSum)
    # print 'tmp',tmpSum
    # print 'max',maxSum
    tmpSum=0
print (maxSum)

15

解答

#改进:如果遇到某个数，比前面的tmpSum+i还大(说明tmpSum<0)，那么可以直接从这里开始
class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        maxSum=nums[0]
        tmpSum=0
        for i in nums:
            tmpSum=max(i,tmpSum+i)
            maxSum=max(maxSum,tmpSum)
        return maxSum

24ms,beat 99.98%

需要注意的点

• 暴力计算不可取，还是要找找规律...这道题就是需要判断符号，假如都是正数，那么肯定是元素越多，加起来越大；而为了简便，其实一旦发现tmpSum变为负，就说明需要重新开始了，那么就从两轮循环变成了一轮循环
• 遍历元素的时候，能不用下标就不用，容易越界