題目：https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/description/

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:

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

二分答案，用lower_bound和upper_bound的思想。

這居然是一道hard題，難以置信，就這個難度啊。

我被自己蠢哭了，把它當成ACM題，以為時間只有一秒，非要優(yōu)化到O(n*log(n))。

沒想到時間和空間復雜度O（n*m）就可以過了。

不說了，第一次過hard題，20分鐘，直接上代碼把：

class Solution { public:int findKthNumber(int m, int n, int k) {if(m > n) swap(m, n);int l = 1; int r = m*n;while(l < r){int mid = (l+r)/2;int p = judge(m, n, k, mid);// cout << p << " ";if(p == 1){return mid;}else if(p == 2){r = mid-1;}else if(p == 3){l = mid+1;}}return r;}int judge(int m, int n, int k, int mid){int sum1 = 0, sum2 = 0;for(int i = 1;i <= m; i++){if(i * n <= mid){sum2 += n;}else{sum2 += mid/i;}if(i * n < mid){sum1 += n;}else{sum1 += (mid-1)/i;}// cout << sum1 << endl;}cout << mid << " " << sum1 << " " << sum2 << endl;if(sum1 < k && sum2 >= k){return 1;}else if(sum1 >= k){return 2;}else if(sum2 < k){return 3;}} }; 與50位技術專家面對面20年技術見證，附贈技術全景圖

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