leetcode378 Kth Smallest Element in a Sorted Matrix

思路1:

使用堆。

实现：

 class Solution
 {
 public:
     int kthSmallest(vector<vector<int>>& matrix, int k)
     {
         using pii = pair<int, int>;
         priority_queue<pii, vector<pii>, greater<pii>> q;
         int n = matrix.size();
         vector<int> v(n, );
         for (int i = ; i < n; i++) q.push(make_pair(matrix[i][v[i]], i));
         pair<int, int> tmp;
         for (int i = ; i < k; i++)
         {
             tmp = q.top(); q.pop();
             int id = tmp.second;
             while (v[id] == n)
             {
                 tmp = q.top();
                 q.pop();
                 id = tmp.second;
             }
             assert(v[id] != n);
             v[id]++;
             q.push(make_pair(matrix[id][v[id]], id));
         }
         return tmp.first;
     }
 };

思路2:

二分查找最小的x，满足大于x的元素数量不超过n * n - k个。

实现：

 class Solution
 {
 public:
     bool check(vector<vector<int>>& matrix, int x, int g)
     {
         int n = matrix.size(), cnt = ;
         for (int i = ; i < n; i++)
         {
             int p = upper_bound(matrix[i].begin(), matrix[i].end(), x) - matrix[i].begin();
             cnt += n - p;
         }
         return cnt <= g;
     }
     int kthSmallest(vector<vector<int>>& matrix, int k)
     {
         int n = matrix.size();
         if (n == ) return matrix[][];
         int l = matrix[][], r = matrix[n - ][n - ];
         int ans = l;
         while (l <= r)
         {
             int m = l + r >> ;
             if (check(matrix, m, n * n - k))
             {
                 ans = m;
                 r = m - ;
             }
             else l = m + ;
         }
         return ans;
     }
 };

思路3:

由于矩阵每一行和每一列都是递增的，可以在思路2的基础上进行优化。

实现：

 class Solution
 {
 public:
     bool check(vector<vector<int>>& matrix, int x, int g)
     {
         int n = matrix.size(), j = n - , cnt = ;
         for (int i = ; i < n; i++)
         {
             while (j >=  && matrix[i][j] > x) j--; //优化，不必每次都二分
             cnt += n -  - j;
         }
         return cnt <= g;
     }
     int kthSmallest(vector<vector<int>>& matrix, int k)
     {
         int n = matrix.size();
         if (n == ) return matrix[][];
         int l = matrix[][], r = matrix[n - ][n - ];
         int ans = l;
         while (l <= r)
         {
             int m = l + r >> ;
             if (check(matrix, m, n * n - k))
             {
                 ans = m;
                 r = m - ;
             }
             else l = m + ;
         }
         return ans;
     }
 };