Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.
Note that it is the kth smallest element in the sorted order, not the kth distinct element.
Example:
matrix = [ [ 1, 5, 9], [10, 11, 13], [12, 13, 15] ], k = 8, return 13.
Note:
You may assume k is always valid, 1 ≤ k ≤ n2.
1 public class Solution { 2 public int kthSmallest(int[][] matrix, int k) { 3 int n = matrix.length; 4 PriorityQueue<Wrapper> pq = new PriorityQueue<Wrapper>(); 5 for(int i=0;i<n;i++) pq.offer(new Wrapper(0,i,matrix[0][i])); 6 for(int i=0;i<k-1;i++){ 7 Wrapper w = pq.poll(); 8 int x = w.x; 9 if(x==n-1) continue; 10 pq.offer(new Wrapper(x+1,w.y,matrix[x+1][w.y])); 11 } 12 return pq.poll().val; 13 } 14 class Wrapper implements Comparable<Wrapper>{ 15 int x; 16 int y; 17 int val; 18 public Wrapper(int x,int y,int val){ 19 this.x = x; 20 this.y = y; 21 this.val = val; 22 } 23 public int compareTo(Wrapper that){ 24 return this.val-that.val; 25 } 26 } 27 } 28 //the run time complexity could be O(klongn), the space complexity could be O(k);
1 public class Solution { 2 public int kthSmallest(int[][] matrix, int k) { 3 int left = matrix[0][0]; 4 int right = matrix[matrix.length-1][matrix.length-1]+1; 5 while(left<right){ 6 int mid = left +(right-left)/2; 7 int count = 0; 8 int j = matrix.length-1; 9 for(int i=0;i<matrix.length;i++){ 10 while(j>=0&&matrix[i][j]>mid){ 11 j--; 12 } 13 count+=j+1; 14 } 15 if(count<k) left = mid+1; 16 else right = mid; 17 } 18 return left; 19 } 20 } 21 //the run time complexity could be O()???