Contiguous Array

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: 2Explanation: [0, 1] is the longest contiguous subarray with equal number of 0 and 1.

Example 2:

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

解析：

开始实在没想到，看到大佬的解法恍然大悟，以后遇到求子数组的问题，要想到累加计数，遇到1时sum++,遇到0时sum--,并且记录每次sum的值与当前累加的位置，下次遇到相同的sum时，与最开始遇到的sum的位置之间的子数组的和即为0.

代码：

class Solution {public:    int findMaxLength(vector<int>& nums) {        int zerosnum=0;        int onesnum=0;        int sum=0;        unordered_map<int,int>m;        m[0]=-1;        int ans=0;        for (int i=0; i<nums.size(); i++)        {            if (nums[i]==1)            sum++;            else            sum--;            if (m.find(sum)!=m.end())            {                ans=max(ans,i-m[sum]);            }            else            {                m[sum]=i;            }        }        return ans;    }};