add: Arithmetic Slices

Author: hijiangtao

README.md

@@ -55,6 +55,7 @@ This is the solutions collection of my LeetCode submissions, most of them are pr
 |371|[Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/) | [JavaScript](./src/sum-of-two-integers/res.js)|Easy|
 |384|[Shuffle an Array](https://leetcode.com/problems/shuffle-an-array/) | [JavaScript](./src/shuffle-an-array/res.js)|Medium|
 |404|[Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves/) | [JavaScript](./src/sum-of-left-leaves/res.js)|Easy|
+|413|[Arithmetic Slices](https://leetcode.com/problems/arithmetic-slices/) | [JavaScript](./src/arithmetic-slices/res.js)|Medium|
 |416|[Partition Equal Subset Sum](https://leetcode.com/problems/partition-equal-subset-sum/) | [JavaScript](./src/partition-equal-subset-sum/res.js)|Medium|
 |434|[Number of Segments in a String](https://leetcode.com/problems/number-of-segments-in-a-string/) | [JavaScript](./src/number-of-segments-in-a-string/res.js)|Easy|
 |494|[Target Sum](https://leetcode.com/problems/target-sum/) | [JavaScript](./src/target-sum/res.js)|Medium|

src/arithmetic-slices/res.js

@@ -0,0 +1,54 @@
+/**
+ * res.js
+ * @authors Joe Jiang ([email protected])
+ * @date    2017-04-13 11:02:22
+ * 
+ * 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.
+ * 
+ * @param {number[]} A
+ * @return {number}
+ */
+let numberOfArithmeticSlices = function(A) {
+	let len = A.length;
+	if (len < 3) {
+		return 0;
+	}
+
+	let res = 0,
+		interval = [];
+
+	// 构建元素差数组
+	for (let i = 1; i < len; i++) {
+		interval.push(A[i] - A[i - 1]);
+	}
+
+	let count = 2,
+		difPre = interval[0];
+
+	len--;
+	for (let i = 1; i < len; i++) {
+		let difNow = interval[i];
+
+		if (difNow === difPre) {
+			count++;
+		} else {
+			res += Sum(count); // 计算当前积攒元素个数下的 arithmetic 个数
+			difPre = difNow;
+			count = 2;
+		}
+	}
+
+	if (count > 2) {
+		res += Sum(count);
+	}
+
+	return res;
+};
+
+let Sum = function(n) {
+	// console.log(n);
+	return (n - 2) * (n - 1) / 2;
+};