[Leetcode]668.Kth Smallest Number in Multiplication Table

链接：LeetCode668

给定高度m 、宽度n 的一张 m * n的乘法表，以及正整数k，你需要返回表中第k 小的数字。

例 1：

输入: m = 3, n = 3, k = 5
输出: 3
解释:
乘法表:
1 2 3
2 4 6
3 6 9

第5小的数字是 3 (1, 2, 2, 3, 3).

相关标签：二分查找

乘法表的特点是每行和每列元素均按升序排序，这题就可以转换为[LeetCode]378.Kth Smallest Element in a Sorted Matrix。我们通过二分查找不断逼近真实值即可。
代码如下：

python:

class Solution:
    def findKthNumber(self, m: int, n: int, k: int) -> int:
        lo, hi = 1, m*n
        while lo <= hi:
            mid = (lo + hi) >> 1
            loc = self.countLower(m,n, mid)
            if loc < k:
                lo = mid + 1
            else:
                hi = mid - 1
        return lo

    def countLower(self, m,n, num):
        i, j = 1,n
        cnt = 0
        while i <= m and j >= 1:
            if i*j <= num:
                cnt += j
                i += 1
            else:
                j -= 1
        return cnt

C++:

class Solution {
public:
    int findKthNumber(int m, int n, int k) {
        int lo=1,hi=m*n;
        while(lo<=hi){
            mid = lo+((hi-lo)>>1);
            cnt = countSamller(m,n,mid);
            if(k <= cnt){
                hi = mid-1;
            } else {
                lo = mid+1;
            }
        }
        return lo;
    }

    int countSamller(int m,int n,int target){
        int i=1,j=n;
        int cnt = 0;
        while(i<=m && j>=1){
            if(target<(i*j)){
                j--;
            }else{
                cnt += j;
                i++;
            }
        }
        return cnt;
    }
};