leetcode413. Arithmetic Slices

摘要： 題目要求 A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elemen...

題目要求

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.

將包含大於等於三個元素且任意相鄰兩個元素之間的差相等的陣列成為等差數列。現在輸入一個隨機陣列，問該陣列中一共可以找出多少組等差數列。

思路一：動態規劃

假設已經知道以第i-1個數字為結尾有k個等差數列，且第i個元素與i-1號元素和i-2號元素構成了等差數列，則第i個數字為結尾的等差數列個數為k+1。因此我們可以自底向上動態規劃，記錄每一位作為結尾的等差數列的個數，並最終得出整個數列中等差數列的個數。程式碼如下：

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

思路二：算數方法

首先看一個簡單的等差數列1 2 3 ， 可知該數列中一共有1個等差數列

再看1 2 3 4 ， 可知該數列中一共有3個等差數列，其中以3為結尾的1個，以4為結尾的2個

再看1 2 3 4 5 , 可知該數列中一共有6個等差數列，其中以3為結尾的1個，4為結尾的2個，5為結尾的3個。

綜上，我們可以得出，如果是一個最大長度為n的等差數列，則該等差數列中一共包含的等差數列個數為(n-2+1)*(n-2)/2 ，即(n-1)*(n-2)/2 。

因此，我們只需要找到以當前起點為開始的最長的等差數列，計算該等差數列的長度並根據其長度得出其共包含多少個子等差數列。

程式碼如下：

public int numberOfArithmeticSlices2(int[] A) {
if(A.length <3) return 0;
int diff = A[1]-A[0];
int left = 0;
int right = 2;
int count = 0;
while(right < A.length) {
if(A[right] - A[right-1] != diff) {
count += (right-left-1) * (right-left-2) / 2;
diff = A[right] - A[right-1];
left = right-1;
}
right++;
}
count += (right-left-1) * (right-left-2) / 2;
return count;
}