# Contiguous Array(2 sum)

Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1.

**Example 1:**

```
Input:
 [0,1]

Output:
 2

Explanation:
 [0, 1] is the longest contiguous subarray with equal number of 0 and 1.
```

**Example 2:**

```
Input:
 [0,1,0]

Output:
 2

Explanation:
 [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
```

**Note:**The length of the given binary array will not exceed 50,000.

分析

two sum的变种

count ++ if 1 count--if 0. count 0，1. map存 count: index，初始{0，-1} 此处-1是Index，为了后面计算正确。可用{0，1}验证

这里ret直接index j-i, 比如{1，0，1} 。因为一样sum又出现，说明中间的array accumulated sum=0 此时尾含头不含，正好j-i。

```
class Solution(object):
    def findMaxLength(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """

        ret, sum2idx, count = 0, {0: -1}, 0
        for i, v in enumerate(nums):
            count += -1 if v == 0 else 1
            if count in sum2idx:
                ret = max(ret, i - sum2idx[count])
            else:
                sum2idx[count] = i

        return ret
```