• [LeetCode][JavaScript]Maximum Gap


    Maximum Gap

    Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

    Try to solve it in linear time/space.

    Return 0 if the array contains less than 2 elements.

    You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

    https://leetcode.com/problems/maximum-gap/


    这道题正确的姿势是先随便写一把快排交过了,然后去看Solution。 233

    原版Solution解释得很清楚了,抽屉原理+箱排序。

    Suppose there are N elements and they range from A to B.

    Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]

    Let the length of a bucket to be len = ceiling[(B - A) / (N - 1)], then we will have at most num = (B - A) / len + 1 of bucket

    for any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K - A) / len and therefore maintain the maximum and minimum elements in each bucket.

    Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken from two elements in the same buckets.

    For each non-empty buckets p, find the next non-empty buckets q, then q.min - p.max could be the potential answer to the question. Return the maximum of all those values.

    Analysis written by @porker2008.

     1 /**
     2  * @param {number[]} nums
     3  * @return {number}
     4  */
     5 var maximumGap = function(nums) {
     6     if(nums.length < 2){
     7         return 0;
     8     }
     9     var boxMap = {};
    10     var max = Math.max.apply(null, nums);
    11     var min = Math.min.apply(null, nums);
    12     var vol = parseInt((max - min) / (nums.length - 1));
    13     if(vol === 0){
    14         vol = 1;
    15     }
    16     
    17     var i = 0, box = null, curr = null;
    18     for(i = 0; i < nums.length; i++){
    19         curr = nums[i];
    20         box = parseInt((curr - min) / vol);
    21         if(!boxMap[box]){
    22             boxMap[box] = {max : curr, min : curr};
    23         }else{
    24             if(curr > boxMap[box].max){
    25                 boxMap[box].max = curr;
    26             }else if(curr < boxMap[box].min){
    27                 boxMap[box].min = curr;
    28             }
    29         }
    30     }
    31     
    32     var maxGap = -Infinity;
    33     var previousMax = null;
    34     for(i in boxMap){
    35         if(previousMax && (boxMap[i].max - previousMax)> maxGap){
    36             maxGap = boxMap[i].min - previousMax;
    37         }
    38         previousMax = boxMap[i].max;
    39     }
    40     return maxGap;
    41 };
  • 相关阅读:
    晒一下我的统一网站管理平台
    走向DBA[MSSQL篇] 从SQL语句的角度 提高数据库的访问性能
    XSS跨站脚本实例
    关于开源的一点看法
    晒一下我的监控系统
    走向DBA[MSSQL篇] 详解游标
    【linux+C】新征程 linux下C编程
    走向DBA[MSSQL篇] 针对大表 设计高效的存储过程【原理篇】 附最差性能sql语句进化过程客串
    晒一下我的web过滤器
    分享新出炉的微软派工具 你,值得拥有
  • 原文地址:https://www.cnblogs.com/Liok3187/p/4603453.html
Copyright © 2020-2023  润新知