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.

思路1：利用一个计数值表示0和1相差多少，然后若在map中找到相应的映射关系，若i和j的cnt相等，表明 i+1 到 j 的0，1个数相等；

int findMaxLength1(vector<int>& nums) {    unordered_map<int, int> res;    int cnt = 0, maxlen = 0;    res[0] = -1;    for (int i = 0; i < nums.size(); i++){        if (!nums[i])cnt++;        else cnt--;        if (res.find(cnt) == res.end())res[cnt] = i;        else maxlen = max(maxlen, i - res[cnt]);    }    return maxlen;}

思路2：将数组中等于0的元素值改为-1，这样若数组中存在连续的两端具有相同的sum，则此区间存在相同数目的0和1。

int findMaxLength(vector<int>& nums) {    for (int i = 0; i < nums.size(); i++){        if (!nums[i])nums[i] = -1;    }    unordered_map<int, int> res;    int sum = 0, maxlen = 0;    res[0] = -1;    for (int i = 0; i < nums.size(); i++){        sum += nums[i];        if (res.find(sum) != res.end()) maxlen = max(maxlen, i - res[sum]);        else res[sum] = i;    }    return maxlen;}