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_632.java
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_632.java
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package com.fishercoder.solutions;
import java.util.List;
import java.util.PriorityQueue;
/**
* 632. Smallest Range
*
* You have k lists of sorted integers in ascending 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:[[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].
Note:
The given list may contain duplicates, so ascending order means >= here.
1 <= k <= 3500
-105 <= value of elements <= 105.
For Java users, please note that the input type has been changed to List<List<Integer>>.
And after you reset the code template, you'll see this point.
*/
public class _632 {
/**
* reference: https://discuss.leetcode.com/topic/94445/java-code-using-priorityqueue-similar-to-merge-k-array/2
*/
public int[] smallestRange(List<List<Integer>> nums) {
PriorityQueue<int[]> minHeap = new PriorityQueue<>(nums.size(), (a, b) -> a[0] - b[0]);
/**int[] array consists of three numbers: value; which list in nums; index of value in this list*/
int max = nums.get(0).get(0);
for (int i = 0; i < nums.size(); i++) {
minHeap.offer(new int[]{nums.get(i).get(0), i, 0});
max = Math.max(max, nums.get(i).get(0));
}
int minRange = Integer.MAX_VALUE;
int start = -1;
while (minHeap.size() == nums.size()) {
int[] curr = minHeap.poll();
if (max - curr[0] < minRange) {
minRange = max - curr[0];
start = curr[0];
}
if (curr[2] + 1 < nums.get(curr[1]).size()) {
curr[0] = nums.get(curr[1]).get(curr[2] + 1);
curr[2]++;
minHeap.offer(curr);
max = Math.max(max, curr[0]);
}
}
return new int[]{start, start + minRange};
}
}