难度：容易

1. Description

138. Subarray Sum

2. Solution

• python
  暴力方法，时间复杂度O(n^2)

class Solution:
    """
    @param nums: A list of integers
    @return: A list of integers includes the index of the first number and the index of the last number
    """
    def subarraySum(self, nums):
        # write your code here
        for i in range(len(nums)):
            m_sum = 0
            for j in range(i,len(nums)):
                m_sum += nums[j]
                if(m_sum==0):
                    return [i,j]

用前缀和，时间复杂度O(n)
前缀和prefix_sum[i] = nums[0]+nums[1]+...nums[i]，即数组前i项和。
利用nums[i+1]+nums[i+1]+...+nums[j] = prefix_sum[j]-prefix_sum[i]，
当prefix_sum[j]=prefix_sum[i]时，i+1到j的子串和为0。

class Solution:
    """
    @param nums: A list of integers
    @return: A list of integers includes the index of the first number and the index of the last number
    """
    def subarraySum(self, nums):
        # write your code here
        prefix_hash = {0:-1}
        prefix_sum = 0
        for i in range(len(nums)):
            prefix_sum += nums[i]
            if prefix_sum in prefix_hash.keys():
                return prefix_hash[prefix_sum]+1,i
            prefix_hash[prefix_sum] = i
        return -1,-1

3. Reference

1. https://www.lintcode.com/problem/subarray-sum/description