【leetcode】668. Kth Smallest Number in Multiplication Table

题目如下：

解题思路：几乎和【leetcode】719. Find K-th Smallest Pair Distance 的方法一样。只不过一个是减法一个是乘法，还有一点区别是【leetcode】719. Find K-th Smallest Pair Distance中i-j和j-i只算一个元素，而本题中i*j与j*i算两个元素。

代码如下：

class Solution(object):
    def findKthNumber(self, m, n, k):
        """
        :type m: int
        :type n: int
        :type k: int
        :rtype: int
        """
        low,high = 1, m*n

        while low <= high:
            mid = (low + high) / 2
            less,equal = 0,0
            for i in range(1,min(mid+1,m+1)):
                less += min(mid/i,n)
                if mid/i <= n and mid % i == 0:
                    equal += 1
                    less -= 1

            if less >= k:
                high = mid - 1
            elif less + equal < k:
                low = mid + 1
            elif less == k and equal == 0:
                high = mid - 1
            else:
                break
        return mid