LeetCode: 413. Arithmetic Slices 数组中等差子序列的个数

分手后的思念是犯贱 2021-06-24 16:10 388阅读 0赞

**试题：**  
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.

**代码：**  
dp\[i\]仅是存储以A\[i\]结尾的等差数列。从dp\[i-1\]到dp\[i\]，一种是dp\[i-1\]已经保存的，另外就是新加一个。  
此处动态规划求解了每个位置结尾的子序列数量，最终结果还需要求和。

class Solution {
        public int numberOfArithmeticSlices(int[] A) {
            if(A==null || A.length==0) return 0;
            int len = A.length;
            int[] dp = new int[len];
            
            for(int i=2; i<len; i++){
                if(A[i]-A[i-1]==A[i-1]-A[i-2]){
                    dp[i] = dp[i-1] + 1;
                }
            }
            
            int total=0;
            for(int num : dp){
                total += num;
            }
            return total;
        }
    }