zoukankan      html  css  js  c++  java
  • LeetCode 446. Arithmetic Slices II

    原题链接在这里:https://leetcode.com/problems/arithmetic-slices-ii-subsequence/

    题目:

    A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

    For example, these are arithmetic sequences:

    1, 3, 5, 7, 9
    7, 7, 7, 7
    3, -1, -5, -9

    The following sequence is not arithmetic.

    1, 1, 2, 5, 7

    A zero-indexed array A consisting of N numbers is given. A subsequence slice of that array is any sequence of integers (P0, P1, ..., Pk) such that 0 ≤ P0 < P1 < ... < Pk < N.

    A subsequence slice (P0, P1, ..., Pk) of array A is called arithmetic if the sequence A[P0], A[P1], ..., A[Pk-1], A[Pk] is arithmetic. In particular, this means that k ≥ 2.

    The function should return the number of arithmetic subsequence slices in the array A.

    The input contains N integers. Every integer is in the range of -231 and 231-1 and 0 ≤ N ≤ 1000. The output is guaranteed to be less than 231-1.

    Example:

    Input: [2, 4, 6, 8, 10]
    
    Output: 7
    
    Explanation:
    All arithmetic subsequence slices are:
    [2,4,6]
    [4,6,8]
    [6,8,10]
    [2,4,6,8]
    [4,6,8,10]
    [2,4,6,8,10]
    [2,6,10]

    题解:

    Following question: Arithmetic Slices.

    The difference here is that subsequence doesn't need to be continuous. e.g. [2,6,10].

    Thus when iterate to index i, it needs to go through all the j (j<i) from the beginning.

    With A[i] and A[j], the diff = A[i] - A[j]. It needs to know the number of sequences(length doesn't need to be more than 2, 2 is okay) ending at A[j] with the same diff. Accumulate that to result.

    Thus here, use an array of map to maintain the difference with frequency for each index.

    There could be duplicate numbers in A. Thus at i, same diff may appear, get the original and add the new.

    Time Complexity: O(n^2). n = A.length.

    Space: O(n^2). Each map could be O(n), there are totally n maps.

    AC Java:

     1 class Solution {
     2     public int numberOfArithmeticSlices(int[] A) {
     3         int n = A.length;
     4         Map<Integer, Integer> [] arrOfMap = new Map[n];
     5         int res = 0;
     6         
     7         for(int i = 0; i<n; i++){
     8             arrOfMap[i] = new HashMap<>();
     9             
    10             for(int j = 0; j<i; j++){
    11                 long diffLong = (long)A[i] - A[j];
    12                 if(diffLong > Integer.MAX_VALUE || diffLong < Integer.MIN_VALUE){
    13                     continue;
    14                 }
    15                 
    16                 int diff = (int)diffLong;
    17                 int count = arrOfMap[j].getOrDefault(diff, 0);
    18                 res += count;
    19                 
    20                 int original = arrOfMap[i].getOrDefault(diff, 0);
    21                 arrOfMap[i].put(diff, original+count+1);
    22             }
    23         }
    24         
    25         return res;
    26     }
    27 }
  • 相关阅读:
    千个常用DOS命令全面收藏
    面向对象设计的11原则
    SQL语句判断指定的数据库、表、字段、存储过程是否存在
    ASP.NET MVC2 Areas区域新概念
    标准的 SQL 解析顺序
    Improvements to workspaces in TFS 2010
    jquery ajax return值不能取得的解决方案
    用 SQL 语句创建数据库用户(SQL Server 2005)
    简单实现.net MVC自定义错误处理页面
    自定义截图类(C#)
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11441949.html
Copyright © 2011-2022 走看看