题目描述：

LeetCode 668. Kth Smallest Number in Multiplication Table

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:

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

题目大意：

求m * n乘法表的第k个数

解题思路：

从1到k进行二分枚举，上下界分别为lo, hi，记当前枚举数字为mid

利用O(n)的代价求乘法表中有多少个数字不大于mid，记为count

  若count >= k，则令hi = mid - 1

  否则，令lo = mid + 1

最后返回lo

Python代码：

class Solution(object):
    def findKthNumber(self, m, n, k):
        """
        :type m: int
        :type n: int
        :type k: int
        :rtype: int
        """
        count = lambda t: sum(min(m, t / x) for x in range(1, n + 1))
        lo, hi = 1, k
        while lo <= hi:
            mid = (lo + hi) / 2
            if (count(mid)) >= k:
                hi = mid - 1
            else:
                lo = mid + 1
        return lo

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