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题目：

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.

解释：

动态规划 解法1： 用DP来做，定义一个一维dp数组，其中dp[i]表示，到i位置为止的Arithmetic Slices的个数，那么我们从第三个数字开始遍历，如果当前数字和之前两个数字构成Arithmetic Slices，那么我们更新dp[i]为dp[i-1]+1 dp[i]=dp[i-1]+1可以理解为，当A[i]可以和A[i-1]以及A[i-2]构成Arithmetic Slices的时候，那么能和A[i-1]构成Arithmetic Slices的元素必定能和A[i]构成算数切片，所以有个dp[i-1]，除此之外，当前这个最新元素还可以和它前面的两个元素构成Arithmetic Slices，而它前面的两个元素在dp[i-1]的时候只有两个元素，无法构成Arithmetic Slices，故+1。 然后res累加上dp[i]的值即可 解法2： 还可以进一步优化空间，由于dp[i]只和dp[i-1]有关，所以可以用一个变量来代替上面的数组，原理都一样。 python代码：

class Solution(object):    def numberOfArithmeticSlices(self, A):        """        :type A: List[int]        :rtype: int        """        #解法1：        len_A=len(A)        if len_A<3:            return 0        dp=[0]*len_A        result=0        for i in xrange(2,len_A):            if 2*A[i-1]==A[i-2]+A[i]:                dp[i]=dp[i-1]+1            result+=dp[i]        return result

class Solution(object):    def numberOfArithmeticSlices(self, A):        """        :type A: List[int]        :rtype: int        """        #解法2：        len_A=len(A)        if len_A<3:            return 0        cur=0        result=0        for i in xrange(2,len_A):            if A[i-1]*2==A[i]+A[i-2]:                cur+=1                result+=cur            else:                cur=0        return result

c++代码：

class Solution {
   public:    int numberOfArithmeticSlices(vector
   
    & A) {
           //解法1        int len_A=A.size();        if (len_A<3)            return 0;        int result=0;        vector
    
     dp(len_A,0);        for (int i =2;i
    
   

class Solution {
   public:    int numberOfArithmeticSlices(vector
   
    & A) {
           //解法2        int len_A=A.size();        if (len_A<3)            return 0;        int result=0;        int cur=0;        for (int i =2;i
   

总结：

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