LeetCode 446. Arithmetic Slices II - Subsequence
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题目要求
A sequence of numbers 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 sequences:
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 subsequence slice of that array is any sequence of integers (P0, P1, …, Pk) such that 0 ≤ P0 < P1 < … < Pk < N.
A subsequence slice (P0, P1, …, Pk) of array A is called arithmetic if the sequence A[P0], A[P1], …, A[Pk-1], A[Pk] is arithmetic. In particular, this means that k ≥ 2.
The function should return the number of arithmetic subsequence slices in the array A.
The input contains N integers. Every integer is in the range of -231 and 231-1 and 0 ≤ N ≤ 1000. The output is guaranteed to be less than 231-1.
Example:
Input: [2, 4, 6, 8, 10]
Output: 7
Explanation:
All arithmetic subsequence slices are:
[2,4,6]
[4,6,8]
[6,8,10]
[2,4,6,8]
[4,6,8,10]
[2,4,6,8,10]
[2,6,10]
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题意解析
这道题的含义就是找出数列中所有3个数以上的等差数列的数量。
解法分析
使用动态规划来解这道题。假设dp[i][j]表示以A[i]结尾的子序列(P0, P1, …, Pi)构成的数列，序列中的元素之差为j。而dp[i][j]=dp[k][j]>0?dp[k][j]+1:1，0<=i
解题代码
public int numberOfArithmeticSlices(int[] A) {
int re = 0;
HashMap> diffMaps = new HashMap<>();
for (int i = 0; i < A.length; i++) {
HashMap diffMap = new HashMap<>();
diffMaps.put(i, diffMap);
int num = A[i];
for (int j = 0; j < i; j++) {
if ((long) num - A[j] > Integer.MAX_VALUE)
continue;
if ((long) num - A[j] < Integer.MIN_VALUE)
continue;
long diff = (long) num - A[j];
int count = diffMaps.get(j).getOrDefault(diff, 0);
diffMaps.get(i).put(diff, diffMaps.get(i).getOrDefault(diff, 0) + count + 1);
re += count;
}
}
return re;
}