378. Kth Smallest Element in a Sorted Matrix

最新推荐文章于 2020-06-29 10:46:52 发布

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最新推荐文章于 2020-06-29 10:46:52 发布

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分类专栏： LeetCode Binary Search Heap 文章标签： LeetCode Binary Search Heap Kth Smallest Element

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本文介绍了一种利用二分查找高效求解矩阵中第K小元素的方法。矩阵为n×n大小，每行每列均按升序排列。通过初始化搜索范围并迭代缩小范围直至找到目标元素。

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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:

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;
    }
}