525. Contiguous Array

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.

方法1: presum + hashmap

思路：

如何在遍历一次的过程中就能知道中间是否包括dqual number的subarray呢？我们需要沿途记录0，1的数量。这个时候，可以通过累计和来区分：遇到1则++，遇到0–。那么如果中间某一段的sum是0，代表0和1数量相等。这样在遇到新的一个累计和时，就在hashmap中查找第一次出现同样sum的位置，[i + 1, j] 之间即满足equal sum。需要注意初始化需要手动推入一个{0: -1}的映射：当[0, j]整个和是0的时候，j + 1需要被计入结果，和j - hash[sum]统一起来即需要设置{0: -1}。

Complexity

Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        unordered_map<int, int> hash;
        hash[0] = -1;
        int presum = 0, res = 0;
        for (int i = 0; i < nums.size(); i++) {
            int num = nums[i];
            presum += num ? 1 : -1;
            if (hash.count(presum)) {
                res = max(res, i - hash[presum]);
            }
            else {
                hash[presum] = i;
            }
        }
        return res;
    }
};