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解法

巧妙地將0轉(zhuǎn)換成-1，利用counter記錄前綴和所在的位置，這樣兩個相同的前綴和之間的長度就是連續(xù)的相同數(shù)據(jù)的0和1.

class Solution {
    public int findMaxLength(int[] nums) {
       int maxLength = 0;
       Map<Integer, Integer> map = new HashMap<>();
       int counter = 0;
       map.put(counter, -1);
       for (int i = 0; i < nums.length; i++) {
        int num = nums[i];
        if (num == 0) {
            counter += -1;
        } else {
            counter += 1;
        }
        if (map.containsKey(counter)) {
            maxLength = Math.max(maxLength, i - map.get(counter));    
        } else {
            map.put(counter, i);
        }
       } 
       return maxLength;                    
    }
}