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.
此题似乎没有特别精妙的解法,discuss区排第一的答案看起来就是brute force而已; 复杂度 O(Matrix.height * (Matrix.width + Matrix.height) )
其他的优化也就是用hash进行空间换时间.
1 class Solution { 2 public int maxEqualRowsAfterFlips(int[][] matrix) { 3 int h=matrix.length; 4 int w=matrix[0].length; 5 int res=1; 6 for(int i=0;i<h;++i) 7 { 8 int tmp=0; 9 int []flip=new int[w]; 10 for(int j=0;j<w;++j) 11 flip[j]=1-matrix[i][j]; //flip保存的是第i行 flip后的结果 12 for(int k=0;k<h;++k) 13 { 14 if(Arrays.equals(matrix[k],matrix[i])||Arrays.equals(matrix[k],flip)) //计算完第j行的flip结果后,把每一行比较,如果等于i或者flip后的结果则命中 15 ++tmp; 16 } 17 res=Math.max(res,tmp); 18 } 19 return res; 20 } 21 }
1072. Flip Columns For Maximum Number of Equal Rows
原文：https://www.cnblogs.com/lychnis/p/11638226.html