zoukankan      html  css  js  c++  java
  • The longest plateau

    Problem:

    Given an array, try to develop an efficient algorithm which can compute the length of the longest plateau. A plateau is a consecutive segment of an array with equal contents. For example, if x[] = {1, 2, 3, 4, 4, 4, 5, 5, 6}, then we have six plateaus which are 1, 2, 3, 4-4-4, 5-5 and 6. And obviously the length of the longest plateaus is 3.

    Analysis:

    Well, a straightforward idea is try to firstly compute all the length of different plateaus from left to right and then select the longest length. The pseudo-code is like this:

    for each element in the array a[]
         if a[i] is equal to a[i-1]
              add 1 to the length for the current plateau
              check whether the current length is the longest one so far
         else
              reset length to 1 // plateau length is at least 1

    Whether we need line 5&6 depends on whether we need to store the length of every plateau. If we just want to calculate the longest length then we can keep the code and use the “length” as a temp variable which is only used inside the loop. On the other hand, if we need to keep track of the length of all plateaus, we need to use an array of “length[]” to store the needed information.

    /*
     * input: an array a[], the length of the array n
     * output: the length of the longest plateau
     */
    int longestPlateau (int a[], int n)
    {
        if (n == 0)
            return 0;
     
        int length = 1;
        int longestLength = 1;
     
        for (int i = 1; i<n; i++)
        {
            if (a[i] == a[i-1])
            {
                length++;
                longestLength = max(longestLength, length);
            }
            else
                length = 1;
        }
        return longestLength;
    }

    Some more:

    What if the given array is sorted (in the increasing order) already?

    Actually if the array is sorted, the algorithm can be much simpler:

    assume the longest length now is L, then we just need to compare a[i] and a[i-L], if they are equal then all the elements between them are also equal (since this is a sorted array!), and we can add 1 to the current longest length. The code looks like this:

    /*
     * input: an sorted array a[] (increasing order), the length of the array n
     * output: the length of the longest plateau
     */
    int longestPlateau (int a[], int n)
    {
        if (n == 0)
            return 0;
     
        int length = 1;
        for (int i = 1; i<n; i++)
        {
            if (a[i] == a[i-length])
                length++;
        }
        return length;
    }
  • 相关阅读:
    filter, map, reduce, zip函数
    schwartzian sort
    各种排序
    MVVM
    js/jquery学习笔记(附百度统计初探??)
    由一句需求引发的mysql崩溃说起
    高效前端优化工具Fiddler入门教程
    由一次很有意思的购物体验想到的
    个人电脑文件目录变更日志小程序
    浅谈COOKIE和SESSION关系和区别等
  • 原文地址:https://www.cnblogs.com/zhangjie/p/3340954.html
Copyright © 2011-2022 走看看