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 = 5Output: Explanation: The Multiplication Table:123246369The 5-th smallest number is 3 (1, 2, 2, 3, 3).

Example 2:

Input: m = 2, n = 3, k = 6Output: Explanation: The Multiplication Table:123246The 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]

思路：直观的想法是优先队列，最坏复杂度n*m*log(m)，其中log(m)是插入删除时更新优先队列的代价（heap深度是m）

一个更快的思路是binary Search，直接BS结果，复杂度log(m*n)

这里的下表控制特别要小心

不过最重要的是：binary Search要怎么想到呢？？？因为你只要求第K个数嘛，并不需要求所有0..K的

/* * since最坏的情况下是需要全部遍历一遍数据，优先队列更新数据还要复杂度 * 更快应该就是logN * 可能就是 binary search * 直接对结果二分，判断是不是满足一定条件即可 */class Solution {    public int findKthNumber(int m, int n, int k) {    int lo = 1, hi = m*n, mid = -1;    while(lo < hi) {// 注意这里要变化    mid = lo+(hi-lo)/2;    int t = getNotLarger(m, n, mid);    if(t == k)hi = mid;// 等于还要往左走到乘法表里面有的（不是往右走，不然就是k+1）    else if(t > k)hi = mid;// 因为乘法表里面有重复的数，比如mid有多个，结果仍然可能是mid    elselo = mid+1;    }    return lo;    }private int getNotLarger(int m, int n, int mid) {int cnt = 0;for(int i=1; i<=m; i++)cnt += Math.min(mid/i, n);return cnt;}}