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
andn
will be in the range [1, 30000]. - The
k
will be in the range [1, m * n]
这道题的思路是这样的:m*n乘法表的数值区间为【1,m*n】,左右边界分别设为left ,right。设mid = left + (right-left)/2;从第一行遍历乘法表,统计出中不大于mid的数的个数,如果小于k说明第k个值在【mid+1,right】区间内,若大于等于则说明第k个数在【left,mid】区间内,不断缩小查找区间,直到left==right;时间复杂度为O(n*logn);
这里有一个关键点:为什么这里要写成①,而不是②
① if(count<k) left = mid+1;
else
{
right = mid;
}
② if(count<k) left = mid+1;
else
{
if(count==k) return mid;
else right = mid-1;
}
因为是遍历整个乘法表,某个数可能有多个,②种写法会漏掉一些情况;
class Solution {
public:
int findKthNumber(int m, int n, int k)
{
int left = 1,right = m*n;
int mid;
while(left < right)
{
mid = left + (right - left)/2;
int count = 0;
int j=n;
for(int i=1;i<=m;i++)
{
while(j>=1 && i*j>mid) j--;
count += j;
}
if(count<k) left = mid+1;
else
{
right = mid;
}
}
return left;
}
};