[LintCode] Kth Smallest Number in Sorted Matrix 有序矩阵中第K小的数字

 

Find the kth smallest number in at row and column sorted matrix.

Have you met this question in a real interview? 

Yes

Example

Given k = 4 and a matrix:

[
  [1 ,5 ,7],
  [3 ,7 ,8],
  [4 ,8 ,9],
]

return 5

Challenge 

O(k log n), n is the maximal number in width and height.

 

LeetCode上的原题，请参见我之前的博客Kth Smallest Element in a Sorted Matrix。

 

解法一：

class Solution {
public:
    /**
     * @param matrix: a matrix of integers
     * @param k: an integer
     * @return: the kth smallest number in the matrix
     */
    int kthSmallest(vector<vector<int> > &matrix, int k) {
        priority_queue<int, vector<int>> q;
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = 0; j < matrix[i].size(); ++j) {
                q.push(matrix[i][j]);
                if (q.size() > k) {
                    q.pop();
                }
            }
        }
        return q.top();
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param matrix: a matrix of integers
     * @param k: an integer
     * @return: the kth smallest number in the matrix
     */
    int kthSmallest(vector<vector<int> > &matrix, int k) {
        int left = matrix[0][0], right = matrix.back().back();
        while (left < right) {
            int mid = left + (right - left) / 2, cnt = 0;
            for (int i = 0; i < matrix.size(); ++i) {
                cnt += upper_bound(matrix[i].begin(), matrix[i].end(), mid) - matrix[i].begin();
            }
            if (cnt < k) left = mid + 1;
            else right = mid;
        }
        return left;
    }
};

 

解法三：

class Solution {
public:
    /**
     * @param matrix: a matrix of integers
     * @param k: an integer
     * @return: the kth smallest number in the matrix
     */
    int kthSmallest(vector<vector<int> > &matrix, int k) {
        int left = matrix[0][0], right = matrix.back().back();
        while (left < right) {
            int mid = left + (right - left) / 2;
            int cnt = search_less_equal(matrix, mid);
            if (cnt < k) left = mid + 1;
            else right = mid;
        }
        return left;
    }
    int search_less_equal(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size(), i = m - 1, j = 0, res = 0;
        while (i >= 0 && j < n) {
            if (matrix[i][j] <= target) {
                res += i + 1;
                ++j;
            } else {
                --i;
            }
        }
        return res;
    }
};