LeetCode hard 668. Kth Smallest Number in Multiplication Table（二分答案，一次过了，好开心，哈哈哈哈）

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

 

668. Kth Smallest Number in Multiplication Table

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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:

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

 

 

二分答案，用lower_bound和upper_bound的思想。

这居然是一道hard题，难以置信，就这个难度啊。

我被自己蠢哭了，把它当成ACM题，以为时间只有一秒，非要优化到O(n*log(n))。

 

没想到时间和空间复杂度O（n*m）就可以过了。

 

不说了，第一次过hard题，20分钟，直接上代码把：

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;
        }
    }
};