Problem-solving ideas:
(1) First horizontally and then vertically, and finally use the priority queue to solve
class Solution {
public:
int kthLargestValue(vector<vector<int>>& matrix, int k) {
priority_queue<int,vector<int>, greater<int>> vec;
for(int i=0;i<matrix.size();i++) {
for(int j=1;j<matrix[i].size();j++) {
matrix[i][j]=matrix[i][j]^matrix[i][j-1];
}
}
for(int i=0;i<matrix[0].size();i++) {
for(int j=1;j<matrix.size();j++) {
matrix[j][i]=matrix[j][i]^matrix[j-1][i];
}
}
for(int i=0;i<matrix.size();i++) {
for(int j=0;j<matrix[i].size();j++) {
if(vec.size()<k) {
vec.push(matrix[i][j]);
} else if(matrix[i][j]>vec.top()) {
vec.push(matrix[i][j]);
vec.pop();
}
}
}
return vec.top();
}
};