Leetcode-668. Kth Smallest Number in Multiplication Table[C++][Java]

目录

一、题目描述

二、解题思路

【C++】

【Java】

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Leetcode-668. Kth Smallest Number in Multiplication Tablehttps://leetcode.com/problems/kth-smallest-number-in-multiplication-table/description/

一、题目描述

Nearly everyone has used the Multiplication Table. The multiplication table of size m x n is an integer matrix mat where mat[i][j] == i * j (1-indexed).

Given three integers m, n, and k, return the kth smallest element in the m x n multiplication table.

Example 1:

Input: m = 3, n = 3, k = 5
Output: 3
Explanation: The 5th smallest number is 3.

Example 2:

Input: m = 2, n = 3, k = 6
Output: 6
Explanation: The 6th smallest number is 6.

Constraints:

• 1 <= m, n <= 3 * 104
• 1 <= k <= m * n

二、解题思路

二分查找

• 总时间复杂度为 O(m * log(m * n))
• 空间复杂度为 O(1)

【C++】

class Solution {
public:
    int findKthNumber(int m, int n, int k) {
        int left = 1, right = m * n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            int count = 0;
            for (int i = 1; i <= m; ++i) {
                count += min(mid / i, n);
            }
            if (count < k) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
};

【Java】

class Solution {
    public int findKthNumber(int m, int n, int k) {
        int left = 1, right = m * n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            int count = 0;
            for (int i = 1; i <= m; i++) {
                count += Math.min(mid / i, n);
            }
            if (count < k) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
}