[Leetcode /] dichotomy matrices of the ordered elements of k small (virtual array half Squeeze +)

Problem Description:

Given a nxn matrix wherein each row and each column of elements in ascending order, find the small element of the matrix k.
Please note that it is the first k small element of a sort, but not the first k elements.

Example:

matrix = [
   [ 1,  5,  9],
   [10, 11, 13],
   [12, 13, 15]
],
k = 8,

Return 13.

Note:
You can assume that the value of k is always valid, 1 ≤ k ≤ n2.

The basic idea:

Here we introduce a virtual array of concepts:

Our dichotomy is not based on our existing array. We just use our existing array of set left and right.

Our mid is based on standards set by our own , he is not necessarily in this array.

AC Code:

class Solution {
public:
    int Count(vector<vector<int>> &m, int target) {
      int count = 0;
      int n = m.size();
      // 利用基本已经排好序的矩阵的性质来统计数据
      for (int i = 0; i < n; ++i) {
        int j = n - 1;
        while (j >= 0) {
          if (m[j][i] <= target) {
            count += j + 1; 
            break;
          } else {
            --j;
          }
        }
      }
      return count;
    }
  
    int kthSmallest(vector<vector<int>>& matrix, int k) {
      int n = matrix.size();    // 这是一个方阵, 所以只要求一个维度就行
      int left = matrix[0][0];
      int right = matrix[n - 1][n - 1];
      // 这里的二分并不是直接基于我们的数据结构
      // 而是我们虚拟化了一个数组
      int mid = left;   // 防止left和right相等的时候，进不去循环
      while (left < right) {
        mid = left + (right - left) / 2;  // mid不再是索引，而是实打实的元素值
        int count = Count(matrix, mid);     // 在数组中寻找比mid小的所有元素个数
        // 夹逼法求得符合要求的元素
        if (count >= k) right = mid;    // 说明mid过大
        else left = mid + 1;
      }
      return left;    // 注意这里是返回left或者right而不是返回mid
    }
};

Other experience:

Note Squeeze binary value is returned when the left and right, rather than the mid value.