zoukankan      html  css  js  c++  java
  • LeetCode 632. Smallest Range Covering Elements from K Lists

    原题链接在这里:https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/

    题目:

    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:

    1. The given list may contain duplicates, so ascending order means >= here.
    2. 1 <= k <= 3500
    3. -105 <= value of elements <= 105.

    题解:

    Consider this question this way, if there are 2 rows, how to do it. We need 2 pointers and move the smaller pointer every time to update smallest range.

    Now we have > 2 rows, we need a PriorityQueue to get the smallest value.

    And we also need to know which row and index of that row, thus we use an array to track all of this.

    Time Complexity: O(n * logk). k = nums.size(). n is length of longest list.

    Space: O(k).

    AC Java: 

     1 class Solution {
     2     public int[] smallestRange(List<List<Integer>> nums) {
     3         if(nums == null || nums.size() == 0){
     4             return new int[0];
     5         }
     6         
     7         int [] res = new int[2];
     8         PriorityQueue<int []> minHeap = new PriorityQueue<>((a, b) -> a[0] - b[0]);
     9         int max = Integer.MIN_VALUE;
    10         int minLen = Integer.MAX_VALUE;
    11         int n = nums.size();
    12         for(int i = 0; i < n; i++){
    13             int num = nums.get(i).get(0);
    14             minHeap.add(new int[]{num, i, 0});
    15             max = Math.max(max, num);
    16         }
    17         
    18         while(minHeap.size() == n){
    19             int [] cur = minHeap.poll();
    20             if(max - cur[0] < minLen){
    21                 minLen = max - cur[0];
    22                 res[0] = cur[0];
    23                 res[1] = max;
    24             }
    25             
    26             if(cur[2] < nums.get(cur[1]).size() - 1){
    27                 int num = nums.get(cur[1]).get(cur[2] + 1);
    28                 minHeap.add(new int[]{num, cur[1], cur[2] + 1});
    29                 max = Math.max(max, num);
    30             }
    31         }
    32         
    33         return res;
    34     }
    35 }
  • 相关阅读:
    常用验证函数isset()/empty()/is_numeric()函数
    jquery select取option的value值发生变化事件
    (转)浅谈HTML5与css3画饼图!
    文本框输入值文字消失常用的两种方法
    简洁的滚动代码(上下滚动)
    (转)PHP的语言结构和函数的区别
    兼容ie7的导航下拉菜单
    jquery中each()函数
    tomcat源码导入eclipse
    weblogic linux环境下新建domain
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12114689.html
Copyright © 2011-2022 走看看