LeetCode 378: Kth Smallest Element in Sorted Matrix
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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; } }
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