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docs/notes/Leetcode 题解 - 动态规划.md
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@ -269,13 +269,35 @@ class NumArray {
[413. Arithmetic Slices (Medium)](https://leetcode.com/problems/arithmetic-slices/description/)
```html
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.
A = [0, 1, 2, 3, 4]
return: 6, for 3 arithmetic slices in A:
[0, 1, 2],
[1, 2, 3],
[0, 1, 2, 3],
[0, 1, 2, 3, 4],
[ 1, 2, 3, 4],
[2, 3, 4]
```
dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
在 A[i] - A[i - 1] == A[i - 1] - A[i - 2] 的条件下，{A[i - 2], A[i - 1], A[i]} 是一个等差递增子区间。如果 {A[i - 3], A[i - 2], A[i - 1]} 是一个等差递增子区间，那么 {A[i - 3], A[i - 2], A[i - 1], A[i]} 也是等差递增子区间，dp[i] = dp[i-1] + 1。
当 A[i] - A[i-1] == A[i-1] - A[i-2]，那么 [A[i-2], A[i-1], A[i]] 构成一个等差递增子区间。而且在以 A[i-1] 为结尾的递增子区间的后面再加上一个 A[i]，一样可以构成新的递增子区间。
```html
dp[2] = 1
[0, 1, 2]
dp[3] = dp[2] + 1 = 2
[0, 1, 2, 3], // [0, 1, 2] 之后加一个 3
[1, 2, 3] // 新的递增子区间
dp[4] = dp[3] + 1 = 3
[0, 1, 2, 3, 4], // [0, 1, 2, 3] 之后加一个 4
[1, 2, 3, 4], // [1, 2, 3] 之后加一个 4
[2, 3, 4] // 新的递增子区间
```
综上，在 A[i] - A[i-1] == A[i-1] - A[i-2] 时，dp[i] = dp[i-1] + 1。
```java
public int numberOfArithmeticSlices(int[] A) {

28
notes/Leetcode 题解 - 动态规划.md
View File

@ -269,13 +269,35 @@ class NumArray {
[413. Arithmetic Slices (Medium)](https://leetcode.com/problems/arithmetic-slices/description/)
```html
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.
A = [0, 1, 2, 3, 4]
return: 6, for 3 arithmetic slices in A:
[0, 1, 2],
[1, 2, 3],
[0, 1, 2, 3],
[0, 1, 2, 3, 4],
[ 1, 2, 3, 4],
[2, 3, 4]
```
dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
在 A[i] - A[i - 1] == A[i - 1] - A[i - 2] 的条件下，{A[i - 2], A[i - 1], A[i]} 是一个等差递增子区间。如果 {A[i - 3], A[i - 2], A[i - 1]} 是一个等差递增子区间，那么 {A[i - 3], A[i - 2], A[i - 1], A[i]} 也是等差递增子区间，dp[i] = dp[i-1] + 1。
当 A[i] - A[i-1] == A[i-1] - A[i-2]，那么 [A[i-2], A[i-1], A[i]] 构成一个等差递增子区间。而且在以 A[i-1] 为结尾的递增子区间的后面再加上一个 A[i]，一样可以构成新的递增子区间。
```html
dp[2] = 1
[0, 1, 2]
dp[3] = dp[2] + 1 = 2
[0, 1, 2, 3], // [0, 1, 2] 之后加一个 3
[1, 2, 3] // 新的递增子区间
dp[4] = dp[3] + 1 = 3
[0, 1, 2, 3, 4], // [0, 1, 2, 3] 之后加一个 4
[1, 2, 3, 4], // [1, 2, 3] 之后加一个 4
[2, 3, 4] // 新的递增子区间
```
综上，在 A[i] - A[i-1] == A[i-1] - A[i-2] 时，dp[i] = dp[i-1] + 1。
```java
public int numberOfArithmeticSlices(int[] A) {