525. Contiguous Array ----------M

 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.

A:

每个位置，计算 count 1 - count0 的差， 并且存入一个map中。 然后每次找之前diff相同的。

class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        int n = nums.size();
        unordered_map<int, int> M;
        M[0]=-1;
        int res = 0;
        int diff = 0; // count of 1 - count of 0
        for(int i =0;i<n; i++){
            diff += nums[i]==1?1:-1;
            auto iter= M.find(diff);
            if( iter != M.end()){
                res = max(res, i - iter->second);
            }else{
                M[diff] = i;
            }
        }
        return res;
    }
};