## [525\. Contiguous Array](https://leetcode.com/problems/contiguous-array/)

Given a binary array `nums`, return _the maximum length of a contiguous subarray with an equal number of _`0`_ and _`1`.

**Example 1:**

**Input:** nums = \[0,1\]
**Output:** 2
**Explanation:** \[0, 1\] is the longest contiguous subarray with an equal number of 0 and 1.

**Example 2:**

**Input:** nums = \[0,1,0\]
**Output:** 2
**Explanation:** \[0, 1\] (or \[1, 0\]) is a longest contiguous subarray with equal number of 0 and 1.

**Constraints:**

- `1 <= nums.length <= 105`
- `nums[i]` is either `0` or `1`.

class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        int res = 0;
        int n = nums.size();
        int sum = 0;
        unordered_map<int, int> m{{0, -1}};
        for (int i = 0; i < n; ++i) {
            // 遇到 1 就 +1，遇到 0 就 -1 
            sum += (nums[i] == 1) ? 1 : -1;
            // 如果加起來的總和不等於 0 的話
            // 目前的長度為目前的 index i 減掉 m[sum]
            if (m.count(sum)) {
                res = max(res, i - m[sum]);
            } else {
                // 如果加起來的總和等於 0 的話，其值為目前的 index
                m[sum] = i;
            }
        }
        // nums = [0,1,0]
        // m = {{-1: 0}, {0: -1}}

        // nums = [0, 0, 0, 1, 1, 1]
        // m = {{-2: 1}, {-1: 0}, {-3: 2}, {0: -1}}
        return res;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::