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

 

 

import java.util.*;
import java.math.*;
import java.lang.*;
import java.io.*;




class Test {
	
	public static void main(String[] args) {
		int[] A = {1,2,3,4,5};
		int num=0;
		Test test = new Test();
		num = test.numberOfArithmeticSlices(A);
		System.out.println(num);
	}
	
	  
	 
	public int numberOfArithmeticSlices(int[] A) {
		int len = A.length;
		if(len<=2) return 0;
		int[] end = new int[len];
		int[] dp = new int[len];
		
		for(int i=2;i<len;i++) {
			if(A[i]-A[i-1]==A[i-1]-A[i-2]) {
				end[i]=end[i-1]+1;
			}else {
				end[i] = 0;
			}
		}
		for(int j=2;j<len;j++) {
			dp[j] = dp[j-1]+end[j];
		}
		return dp[len-1];
		
	}
	    	
}

 

 

 

思路分析：

 

 

 

      题目意思就是找出所有等差数列的个数。end[i]表示的是包含A[i]的等差数列的个数。举个例子： 1，2，3 和 1，2，3，4；3结尾和4结尾的差别就是1，2，3，4 里面有2，3，4整个等差数列。其他3有的，4都有。所以两个差1。利用遍历的顺序性，我们可以把代码压缩到九行。

 

 

import java.util.*;
import java.math.*;
import java.lang.*;
import java.io.*;

class Test {
	
	public static void main(String[] args) {
		int[] A = {1,2,3,4,5};
		int num=0;
		Test test = new Test();
		num = test.numberOfArithmeticSlices(A);
		System.out.println(num);
	}
	
	  
	 
	public int numberOfArithmeticSlices(int[] A) {
		if(A.length<=2) return 0;
		int sum = 0;
		int end = 0;
		for(int i=2;i<A.length;i++) {
			if(A[i]-A[i-1]==A[i-1]-A[i-2]) {
				end+=1;
				sum+=end;
			}else
				end = 0;
	}
		return sum;
}
}