1072. Flip Columns For Maximum Number of Equal Rows
Medium
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 }