378. Kth Smallest Element in a Sorted Matrix
2017-05-07 08:57
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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:
Note:
You may assume k is always valid, 1 ≤ k ≤ n2.
这道题二分查找解题。这里需要找kth smallest,可以初始化left = matrix[0][0], right = matrix[n - 1][n - 1],对每一行再进行二分查找mid的上限,统计比mid小或者相同的数。当count小于k时,left = mid + 1,否则right = mid。代码如下:
public class Solution {
public int kthSmallest(int[][] matrix, int k) {
int n = matrix.length;
int left = matrix[0][0], right = matrix[n - 1][n - 1];
while (left < right) {
int mid = left + (right - left) / 2;
int count = 0;
for (int i = 0; i < matrix.length; i ++) {
count += upperBound(matrix[i], mid);
}
if (count < k) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
private int upperBound(int[] nums, int target) {
int left = -1, right = nums.length;
while (left + 1 < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > target) {
right = mid;
} else {
left = mid;
}
}
return right;
}
}
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.
这道题二分查找解题。这里需要找kth smallest,可以初始化left = matrix[0][0], right = matrix[n - 1][n - 1],对每一行再进行二分查找mid的上限,统计比mid小或者相同的数。当count小于k时,left = mid + 1,否则right = mid。代码如下:
public class Solution {
public int kthSmallest(int[][] matrix, int k) {
int n = matrix.length;
int left = matrix[0][0], right = matrix[n - 1][n - 1];
while (left < right) {
int mid = left + (right - left) / 2;
int count = 0;
for (int i = 0; i < matrix.length; i ++) {
count += upperBound(matrix[i], mid);
}
if (count < k) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
private int upperBound(int[] nums, int target) {
int left = -1, right = nums.length;
while (left + 1 < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > target) {
right = mid;
} else {
left = mid;
}
}
return right;
}
}
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