直观做法我用了 前缀和然后搜索(i,j) 看是否一样多，复杂度 $O(n^2)$ 显示超时

class Solution:
    def findMaxLength(self, nums: List[int]) -> int:
        n = len(nums)
        t = [0 for i in range(n+1)]
        for i in range(0,n):
            t[i+1] = t[i] + nums[i]
        k = min(n,n-t[n])
        for l in range(2*k,-1,-2):
            for i in range(0,n-l+1):
                if (t[l+i]-t[i])*2 == l:
                    return l
        return 0

同样的前缀和思路 加一个 dictionary记载第一次出现个同一个0 1间个数差的位置 复杂度$O(n)$

class Solution:
    def findMaxLength(self, nums: List[int]) -> int:
        d = {0:-1}
        t, ans = 0, 0
        for i in range(len(nums)):
            t = t + 1 if nums[i] == 0 else t - 1
            if t not in d:
                d[t] = i
            else:
                ans = max(ans, i - d[t])
        return ans