LeetCode每日一题(632. Smallest Range Covering Elements from K Lists)

You have k lists of sorted integers in non-decreasing order. Find the smallest range that includes at least one number from each of the k lists.

We define the range [a, b] is smaller than range [c, d] if b - a < d - c or a < c if b - a == d - c.

Example 1:

Input: nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]]
Output: [20,24]

Explanation:
List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
List 2: [0, 9, 12, 20], 20 is in range [20,24].
List 3: [5, 18, 22, 30], 22 is in range [20,24].

Example 2:

Input: nums = [[1,2,3],[1,2,3],[1,2,3]]
Output: [1,1]

Constraints:

• nums.length == k
• 1 <= k <= 3500
• 1 <= nums[i].length <= 50
• -105 <= nums[i][j] <= 105
• nums[i] is sorted in non-decreasing order.

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整体来说是用 merge sort 的方式

用一个 min heap 来维护最终答案中所有的数字， 每次从 heap 中 pop 出最小值， 将该数字所在 list 中的下一个数字 push 进 heap 中， 此时 range 中的最小值是 heap 中 top 的值， 最大值有可能是刚刚 push 进去的那个值， 我们只需要用原有的最大值与该值进行比较即可。

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use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    pub fn smallest_range(nums: Vec<Vec<i32>>) -> Vec<i32> {
        let mut ans;
        let mut min = i32::MAX;
        let mut max = i32::MIN;
        let mut heap = BinaryHeap::new();
        let mut pointers = vec![0; nums.len()];
        for (i, row) in nums.iter().enumerate() {
            min = min.min(row[0]);
            max = max.max(row[0]);
            heap.push(Reverse((row[0], i)));
        }
        ans = vec![min, max];
        loop {
            let Reverse((_, i)) = heap.pop().unwrap();
            pointers[i] += 1;
            if pointers[i] == nums[i].len() {
                break;
            }
            heap.push(Reverse((nums[i][pointers[i]], i)));
            min = heap.peek().unwrap().0 .0;
            max = max.max(nums[i][pointers[i]]);
            if max - min < ans[1] - ans[0] {
                ans = vec![min, max];
            }
        }
        ans
    }
}