[Leetcode] 525. Contiguous Array 解题报告

题目：

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.

Note: The length of the given binary array will not exceed 50,000.

思路：

总体思路是：我们遍历nums，然后检查截止每个位置的时候，0比1多几个。假设在第i个位置上发现了0比1多x个，在j (j > i)这个位置上又发现0比1多x个，那么就说明在区间[i + 1, j]内，0和1的个数是相同的。为了便于统计，我们可以设一个哈希表，记录从0比1多的个数到最早出现多这么个的位置的映射。为了统一处理各种情况，可以设定hash[0] = -1。代码的空间复杂度是O(n)，时间复杂度是O(n)。

代码：

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