# LeetCode-1027. Longest Arithmetic Sequence

## LeetCode-1027. Longest Arithmetic Sequence

Posted by Jae on July 2, 2019

### 1、题目

Given an array A of integers, return the length of the longest arithmetic subsequence in A.

Recall that a subsequence of A is a list A[i_1], A[i_2], …, A[i_k] with 0 <= i_1 < i_2 < … < i_k <= A.length - 1, and that a sequence B is arithmetic if B[i+1] - B[i] are all the same value (for 0 <= i < B.length - 1).

Example 1:

Input: [3,6,9,12]
Output: 4
Explanation:
The whole array is an arithmetic sequence with steps of length = 3.
Example 2:

Input: [9,4,7,2,10]
Output: 3
Explanation:
The longest arithmetic subsequence is [4,7,10].
Example 3:

Input: [20,1,15,3,10,5,8]
Output: 4
Explanation:
The longest arithmetic subsequence is [20,15,10,5].


Note:

2 <= A.length <= 2000
0 <= A[i] <= 10000


### 2、思路

#### 2.1、确定状态

a[k]-a[k-1] = a[k-1]-a[k-2]


f[i][j]=以元素a[i]结尾且以差值j为标准的的最长算术子序列长度

#### 2.2、转移方程

f[i][j] = 2
f[i][j] = max{f[i-1][j], f[i-2][j],...f[0][j]}


#### 2.4、 计算顺序

f[1][diff], f[2][diff], f[3][diff]...f[n-1][diff]

### 3、实现

public int longestArithSeqLength(int[] A) {
int n = A.length;
int[][] f = new int[n][20000+1];
int max = 2;
for (int i = 1; i < n; i++){
for (int j = 0; j < i; j++){
int diff = A[i]-A[j]+10000;
// 说明前面没有差值相同的元素
if (f[j][diff] == 0){
f[i][diff] = 2;
continue;
}

f[i][diff] = Math.max(f[i][diff], f[j][diff]+1);
max = Math.max(max, f[i][diff]);
}
}
return max;
}


### 4、优化

public int longestArithSeqLength(int[] A) {
int n = A.length;
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;

for (int a : A){
if (a > max) max = a;
if (a < min) min = a;
}

int maxDiff = max - min;
int[][] f = new int[n][maxDiff*2+1];
int ans = 2;
for (int i = 1; i < n; i++){
for (int j = 0; j < i; j++){
int diff = A[i] - A[j] + maxDiff;
if (f[j][diff] == 0){
f[i][diff] = 2;
}
f[i][diff] = Math.max(f[j][diff] + 1, f[i][diff]);
if (f[i][diff] > ans){
ans = f[i][diff];
}
}
}
return ans;
}