leetcode 668. Kth Smallest Number in Multiplication Table 有序矩阵搜索 + 右上角二分搜索

Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly from the multiplication table?

Given the height m and the length n of a m * n Multiplication Table, and a positive integer k, you need to return the k-th smallest number in this table.

Example 1:

Input: m = 3, n = 3, k = 5

Output:

Explanation:

The Multiplication Table:

1 2 3

2 4 6

3 6 9

The 5-th smallest number is 3 (1, 2, 2, 3, 3).

Example 2:

Input: m = 2, n = 3, k = 6

Output:

Explanation:

The Multiplication Table:

1 2 3

2 4 6

The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6).

Note:

The m and n will be in the range [1, 30000].

The k will be in the range [1, m * n]

本题题意就是在一个m*n的乘法表，然后要求找到第k小的数字

建议和leetcode 240. Search a 2D Matrix II 矩阵搜索 + 右上角搜索 和 leetcode 378. Kth Smallest Element in a Sorted Matrix一起学习

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>
#include <cstdlib>
#include <ctime>

using namespace std;

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>
#include <cstdlib>
#include <ctime>

using namespace std;

class Solution
{
public:
int findKthNumber(int m, int n, int k)
{
int left = 1, right = m*n;
while (left < right)
{
int mid = (right - left) / 2 + left;
int count = calaCount(m, n, mid);
if (count < k)
left = mid + 1;
else
right = mid;
}
return left;
}

int calaCount(int row, int col, int target)
{
int res = 0;
int i = 1, j = col;
while (i <= row && j >= 1)
{
if (i*j <= target)
{
res += j;
i++;
}
else
j--;
}
return res;
}
};