zoukankan      html  css  js  c++  java
  • 413. Arithmetic Slices

    A sequence of number 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 sequence:

    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 slice of that array is any pair of integers (P, Q) such that 0 <= P < Q < N.

    A slice (P, Q) of array A is called arithmetic if the sequence:
    A[P], A[p + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.

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

    Example:

    A = [1, 2, 3, 4]
    
    return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.

    Approach #1: 

    class Solution {
    public:
        int numberOfArithmeticSlices(vector<int>& A) {
            int size = A.size();
            int ans = 0;
            if (size < 3) return ans;
            vector<int> B(size-1, 0);
            
            for (int i = 1; i < size; ++i) 
                B[i-1] = A[i] - A[i-1];
            
            for (int i = 0; i < B.size(); ) {
                int j = i;
                while (j + 1 < B.size() && B[i] == B[j+1]) ++j;
                int k = j - i + 1;
                if (k >= 2) ans += k * (k - 1) / 2;
                i = j + 1;
            }
            
            return ans;
        }
    };
    

      

    Approach #2: DP. 

    public class Solution {
        public int numberOfArithmeticSlices(int[] A) {
            int[] dp = new int[A.length];
            int sum = 0;
            for (int i = 2; i < dp.length; i++) {
                if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
                    dp[i] = 1 + dp[i - 1];
                    sum += dp[i];
                }
            }
            return sum;
        }
    }
    

      

    Approach #3: constant space DP. 

    public class Solution {
        public int numberOfArithmeticSlices(int[] A) {
            int dp = 0;
            int sum = 0;
            for (int i = 2; i < A.length; i++) {
                if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
                    dp = 1 + dp;
                    sum += dp;
                } else
                    dp = 0;
            }
            return sum;
        }
    }
    

      

    Approach #4: Formula.

    public class Solution {
        public int numberOfArithmeticSlices(int[] A) {
            int count = 0;
            int sum = 0;
            for (int i = 2; i < A.length; i++) {
                if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
                    count++;
                } else {
                    sum += (count + 1) * (count) / 2;
                    count = 0;
                }
            }
            return sum += count * (count + 1) / 2;
        }
    }
    

      

    Analysis:

    https://leetcode.com/problems/arithmetic-slices/solution/

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    "V租房"搭建微信租房平台,让租房人发起求租需求并接收合适房源回复,提高租房效率 | 36氪
    金融街
    Jsensation | 氪加
    Polyvore
    周翔宇_百度百科
    鸵鸟心态
    新闻:型牌男装:网上订服装,如何将返修率降到5个点以下 | IT桔子
    【案例】舒邑:一个女装品牌的奇葩打法-@i黑马
    专访OPPO李紫贵:ColorOS用户过千万 软硬融合生态版图初现
    关于我们-EIBOA易博男装-互联网品质男装品牌-在线销售男士西服,衬衫,外套,西裤,领带|全场免运费,30天退换货保障
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10403924.html
Copyright © 2011-2022 走看看