LeetCode 378: Kth Smallest Element in Sorted Matrix

1. Use priority queue. Need to check whether one element has been double counted:

class Solution {
public int kthSmallest(int[][] matrix, int k) {
if (matrix.length == 0 || matrix[0].length == 0) return 0;
PriorityQueue<int[]> queue = new PriorityQueue<>((n1, n2) -> n1[0] - n2[0]);
queue.offer(new int[]{matrix[0][0], 0, 0});
int i = 1;
int[] current = new int[3];
Set<Integer> visited = new HashSet<>();
while (!queue.isEmpty() && i++ <= k) {
current = queue.poll();
if (current[1] < matrix.length - 1 && !visited.contains((current[1] + 1)*matrix[0].length + current[2])) {
queue.offer(new int[] {matrix[current[1] + 1][current[2]], current[1] + 1, current[2]});
visited.add((current[1] + 1)*matrix[0].length + current[2]);
}
if (current[2] < matrix[0].length - 1 && !visited.contains(current[1]*matrix[0].length + current[2] + 1)) {
queue.offer(new int[] {matrix[current[1]][current[2] + 1], current[1], current[2] + 1});
visited.add(current[1] * matrix[0].length + current[2] + 1);
}
}
return current[0];
}
}

2 Binary search: For this kind of matrix, binary search should work as counting how many number satify the condition for each column and row.

class Solution {
public int kthSmallest(int[][] matrix, int k) {
if (matrix.length == 0 || matrix[0].length == 0) return 0;
int start = matrix[0][0], end = matrix[matrix.length - 1][matrix[0].length - 1];
while (start < end) {
int mid = start + (end - start) / 2;
int count = 0, j = matrix[0].length - 1;
for (int i = 0; i < matrix.length; i++) {
while (j >= 0 && matrix[i][j] > mid) j--;
count += j + 1;
}

if (count < k) start = mid + 1;
else end = mid;
}
return start;
}
}