zoukankan      html  css  js  c++  java
  • (Java) LeetCode 413. Arithmetic Slices —— 等差数列划分

    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.
    

    想了半天这道题竟然是一道找规律问题…有的时候真的难以捉摸什么问题需要推导,什么问题是找规律。

    如果数组是{1,2,3},结果是1;

    如果数组是{1,2,3,4},结果是3;

    如果数组是{1,2,3,4,5},结果是6;

    如果是数组是{1,2,3,4,5,6},结果是10……

    也就是说如果新扫描到的元素仍能和之前的元素保持等差数列关系,那么新形成的子等差数列个数按照1,2,3,4……的规律增加。其实找到规律这道题就做出来了。所以本题根本是寻找差子数列的终点,并以此时的结果增量更新最终结果。不满足等差数列的时候增量就要重置为1。见下文代码。


    Java

    class Solution {
        public int numberOfArithmeticSlices(int[] A) {
            if (A == null || A.length <= 2) return 0;
            int cur = 1, res = 0, diff = A[1] - A[0];
            for (int i = 2; i < A.length; i++) {
                if (A[i] - A[i-1] == diff) res += cur++;
                else {
                    diff = A[i] - A[i-1];
                    cur = 1;
                }
            }
            return res;
        }   
    }
  • 相关阅读:
    二分LIS模板
    NYOJ16 矩形嵌套 【DAG上的DP/LIS】
    动态规划题库
    洛谷 P1616 疯狂的采药【裸完全背包】
    洛谷 P1049 装箱问题【正难则反/01背包】
    洛谷 P1048 采药【裸01背包】
    洛谷 P1064 金明的预算方案【有依赖的分组背包】
    洛谷 P1064 金明的预算方案【DP/01背包-方案数】
    洛谷 P1060 开心的金明【DP/01背包】
    51nod 1202 不同子序列个数 [计数DP]
  • 原文地址:https://www.cnblogs.com/tengdai/p/9300899.html
Copyright © 2011-2022 走看看