[Swift]LeetCode891. 子序列宽度之和 | Sum of Subsequence Widths

Given an array of integers A, consider all non-empty subsequences of A.

For any sequence S, let the width of S be the difference between the maximum and minimum element of S.

Return the sum of the widths of all subsequences of A. 

As the answer may be very large, return the answer modulo 10^9 + 7. 

Example 1:

Input: [2,1,3]
Output: 6
Explanation:
Subsequences are [1], [2], [3], [2,1], [2,3], [1,3], [2,1,3].
The corresponding widths are 0, 0, 0, 1, 1, 2, 2.
The sum of these widths is 6.

Note:

• 1 <= A.length <= 20000
• 1 <= A[i] <= 20000

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给定一个整数数组 A ，考虑 A 的所有非空子序列。

对于任意序列 S ，设 S 的宽度是 S 的最大元素和最小元素的差。

返回 A 的所有子序列的宽度之和。

由于答案可能非常大，请返回答案模 10^9+7。 

示例：

输入：[2,1,3]
输出：6
解释：
子序列为 [1]，[2]，[3]，[2,1]，[2,3]，[1,3]，[2,1,3] 。
相应的宽度是 0，0，0，1，1，2，2 。
这些宽度之和是 6 。 

提示：

• 1 <= A.length <= 20000
• 1 <= A[i] <= 20000

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Runtime: 380 ms

Memory Usage: 19.3 MB

 1 class Solution {
 2     func sumSubseqWidths(_ A: [Int]) -> Int {
 3         var A = A.sorted(by:<)
 4         var c:Int = 1
 5         var res:Int = 0
 6         var mod:Int = Int(1e9 + 7)
 7         let countA:Int = A.count
 8         var i:Int = 0
 9         while(i < countA)
10         {
11             res = (res + A[i] * c - A[countA - i - 1] * c) % mod
12             i += 1
13             c = (c << 1) % mod
14         }
15         return Int((res + mod) % mod)
16     }
17 }