525. Contiguous Array

class Solution {
    public int findMaxLength(int[] nums) {
        int preSum = 0, maxLength = 0;
        Map<Integer, Integer> preSumToLastIndexMap = new HashMap<>();
        preSumToLastIndexMap.put(0, -1);
        for (int i = 0; i < nums.length; i++) {
            int num = nums[i];
            preSum += num == 0 ? -1 : 1;

            // 当 preSum 再次出现时，说明出现了次数相等的 0 和 1
            Integer lastIndex = preSumToLastIndexMap.get(preSum);
            if (lastIndex == null) {
                preSumToLastIndexMap.put(preSum, i);
            } else {
                maxLength = Math.max(maxLength, i - lastIndex);
            }
        }

        return maxLength;
    }
}

Reference

525. Contiguous Array
剑指 Offer II 011. 0 和 1 个数相同的子数组