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leetcode413. Arithmetic Slices

piglei / 3552人閱讀

摘要：題目要求將包含大于等于三個元素且任意相鄰兩個元素之間的差相等的數(shù)組成為等差數(shù)列?，F(xiàn)在輸入一個隨機數(shù)組，問該數(shù)組中一共可以找出多少組等差數(shù)列。

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.

將包含大于等于三個元素且任意相鄰兩個元素之間的差相等的數(shù)組成為等差數(shù)列。現(xiàn)在輸入一個隨機數(shù)組，問該數(shù)組中一共可以找出多少組等差數(shù)列。

思路一：動態(tài)規(guī)劃

假設(shè)已經(jīng)知道以第i-1個數(shù)字為結(jié)尾有k個等差數(shù)列，且第i個元素與i-1號元素和i-2號元素構(gòu)成了等差數(shù)列，則第i個數(shù)字為結(jié)尾的等差數(shù)列個數(shù)為k+1。因此我們可以自底向上動態(tài)規(guī)劃，記錄每一位作為結(jié)尾的等差數(shù)列的個數(shù)，并最終得出整個數(shù)列中等差數(shù)列的個數(shù)。代碼如下：

    public int numberOfArithmeticSlices(int[] A) {
        int[] dp = new int[A.length];
        int count = 0;
        for(int i = 2 ; i
思路二：算數(shù)方法

首先看一個簡單的等差數(shù)列1 2 3， 可知該數(shù)列中一共有1個等差數(shù)列
再看1 2 3 4， 可知該數(shù)列中一共有3個等差數(shù)列，其中以3為結(jié)尾的1個，以4為結(jié)尾的2個
再看1 2 3 4 5, 可知該數(shù)列中一共有6個等差數(shù)列，其中以3為結(jié)尾的1個，4為結(jié)尾的2個，5為結(jié)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(shù)列，則該等差數(shù)列中一共包含的等差數(shù)列個數(shù)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(shù)列，計算該等差數(shù)列的長度并根據(jù)其長度得出其共包含多少個子等差數(shù)列。


代碼如下：


    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;
    }

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