至少有三个数，任意两个之间的差值相等 Arithmetic Slices

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最新推荐文章于 2022-04-21 18:22:02 发布

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文章标签： python c# 数据结构与算法

原文链接：https://my.oschina.net/liyurong/blog/1601357

本文介绍了一种算法解决方案，用于确定一个整数数组中等差数列片段的数量。等差数列是指序列中任意两个连续元素之间的差保持一致。文章提供了两种实现方式：一种利用动态规划的方法，另一种则通过一个变量来跟踪局部计数。

2019独角兽企业重金招聘Python工程师标准>>>

问题：

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.

解决：

① 动态规划，dp[i]表示当前满足arithmetic slices的三个连续的值的个数，而总的个数是满足条件的个数的和。

class Solution {//2ms
public int numberOfArithmeticSlices(int[] A) {
if (A.length < 3) return 0;
int p = 2;
int len = A.length;
int[] dp = new int[len];
int res = 0;
while(p < len){
if (A[p] - A[p - 1] == A[p - 1] - A[p - 2]){//只要差值相等即可，而不是差值的绝对值
dp[p] = dp[p - 1] + 1;
}
res += dp[p];
p ++;
}
return res;
}
}

② 只使用一个一个变量记录局部个数。

class Solution { //2ms
public int numberOfArithmeticSlices(int[] A) {
if (A.length < 3) return 0;
int p = 2;
int len = A.length;
int res = 0;
int cur = 0;
while(p < len){
if (A[p] - A[p - 1] == A[p - 1] - A[p - 2]){
cur += 1;
res += cur;
}else {
cur = 0;
}
p ++;
}
return res;
}
}