Given a binary array nums, return the maximum length of a contiguous subarray with an equal number of0and1.
Example 1:
**Input:** nums = [0,1]
**Output:** 2
**Explanation:** [0, 1] is the longest contiguous subarray with an equal number of 0 and 1.
Example 2:
**Input:** nums = [0,1,0]
**Output:** 2
**Explanation:** [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
Constraints:
1 <= nums.length <= 105
nums[i] is either 0 or 1.
ac
class Solution {
public int findMaxLength(int[] nums) {
// edge cases
if (nums == null || nums.length == 0) return 0;
Map<Integer, Integer> map = new HashMap<>();
map.put(0, -1);
int sum = 0, max = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i] == 1 ? 1 : -1;
if (map.containsKey(sum)) {
max = Math.max(max, i - map.get(sum));
} else {
map.put(sum, i);
}
}
return max;
}
}
/*
1) maintain sum, meet 1 +1, meet 0 -1; 2) if current sum appear before, then the gap is the subarray we want, calculate the length; 3) put new sum-value in map;
*/