题目如下：
matrixReturn the maximum number of rows that have all values equal after some number of flips.
Example 1:
Input: [[0,1],[1,1]] Output: 1 Explanation: After flipping no values, 1 row has all values equal.Example 2:
Input: [[0,1],[1,0]] Output: 2 Explanation: After flipping values in the first column, both rows have equal values.Example 3:
Input: [[0,0,0],[0,0,1],[1,1,0]] Output: 2 Explanation: After flipping values in the first two columns, the last two rows have equal values.Note:
1 <= matrix.length <= 3001 <= matrix[i].length <= 300matrix[i].length's are equalmatrix[i][j]01
解题思路：把matrix任意一行的的所有元素拼成一个字符串，例如0010110，要把这行变成全是0或者全是1，那么要经过4次或者3次的列变换。变换之后，很显然matrix中字符串为0010110或者1101001的行最后也会变成全为0或者全为1。因此题目就变成了找出matrix中的某一行，使得在整个matrix中和这行相等的行或者相反的行的最多(即0对应1，1对应0的行)。怎么求出最大值的呢？并查集很适合这个场景。
代码如下：
class Solution(object): def maxEqualRowsAfterFlips(self, matrix): """ :type matrix: List[List[int]] :rtype: int """ parent = [i for i in range(len(matrix))] def find(v1): p1 = parent[v1] if p1 != v1: return find(p1) return p1 def union(v1,v2): p1 = find(v1) p2 = find(v2) if p1 <= p2: parent[v2] = p1 else: parent[v1] = p2 def toString(l1): new_l1 = map(lambda x: str(x), l1) return ''.join(new_l1) row = [] row_inverse = [] for i in range(len(matrix)): row.append(toString(matrix[i])) v1_inverse = '' for k in row[i]: v1_inverse += '0' if k == '1' else '1' row_inverse.append(v1_inverse) for i in range(len(matrix)): v1 = row[i] v1_inverse = row_inverse[i] for j in range(i+1,len(matrix)): v2 = row[j] if v1 == v2 or v1_inverse == v2: union(i,j) dic = {} res = 0 for i in range(len(matrix)): p = find(i) dic[p] = dic.setdefault(p,0) + 1 res = max(res,dic[p]) return res