【leetcode】1072. Flip Columns For Maximum Number of Equal Rows

题目如下：

Given a matrix consisting of 0s and 1s, we may choose any number of columns in the matrix and flip every cell in that column.  Flipping a cell changes the value of that cell from 0 to 1 or from 1 to 0.

Return 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. 1 <= matrix.length <= 300
2. 1 <= matrix[i].length <= 300
3. All matrix[i].length‘s are equal
4. matrix[i][j] is 0 or 1

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

原文地址：https://www.cnblogs.com/seyjs/p/11026157.html