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【LeetCode】378. Kth Smallest Element in a Sorted Matrix

程序员文章站 2022-06-16 23:41:27
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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.

题解:最简单的使用最大堆遍历一遍,代码如下:

class Solution {
public:
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        priority_queue<int> q;
        for(int i=0;i<matrix.size();i++){
            for(int j=0;j<matrix[0].size();j++){
                q.push(matrix[i][j]);
                if(q.size()>k) q.pop();
            }
        }
        return q.top();
    }
};

二分法:可以用最大最小值的二分法来求,最小值和最大值在边角,然后找出满足第k小条件的最终解,可以实现一个check方法来判断是第几个,如果小于k说明应该从mid到r开始二分,check方法用来计数,实现计数时可以从左下到右上顺序遍历,代码:

class Solution {
public:
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        int n=matrix.size(),m=matrix[0].size();
        int l=matrix[0][0],r=matrix[n-1][m-1];
        while(l<r){
            int mid=(l+r)/2;
            if(check(matrix,k,mid)) r=mid;
            else l=mid+1;
        }
        return l;
    }
    bool check(vector<vector<int>>& matrix,int k,int mid){
        int n=matrix.size(),m=matrix[0].size(),i=n-1,j=0;
        int cnt=0;
        while(i>=0&&j<m){
            if(matrix[i][j]<=mid)cnt+=i+1,j++;
            else i--;
        }
        if(cnt>=k)return true;
        else return false;
    }
};