给定两个稀疏向量，计算它们的点积（数量积）。

实现类 SparseVector：

SparseVector(nums) 以向量 nums 初始化对象。 dotProduct(vec) 计算此向量与 vec 的点积。 稀疏向量 是指绝大多数分量为 0 的向量。你需要 高效 地存储这个向量，并计算两个稀疏向量的点积。

进阶：当其中只有一个向量是稀疏向量时，你该如何解决此问题？

示例 1：

输入：nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0]

输出：8

解释：v1 = SparseVector(nums1) , v2 = SparseVector(nums2)

v1.dotProduct(v2) = 10 + 03 + 00 + 24 + 3*0 = 8

示例 2：

输入：nums1 = [0,1,0,0,0], nums2 = [0,0,0,0,2]

输出：0

解释：v1 = SparseVector(nums1) , v2 = SparseVector(nums2)

v1.dotProduct(v2) = 00 + 10 + 00 + 00 + 0*2 = 0

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

Example 1:

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Input: x = 2.00000, n = 10
Output: 1024.00000

Example 2:

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Input: x = 2.10000, n = 3
Output: 9.26100

Example 3:

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Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

Constraints:

• -100.0 < x < 100.0
• -231 <= n <= 231-1
• -104 <= xn <= 104

思路

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

A substring is a contiguous sequence of characters within the string.

Example 1:

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Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Given a characters array tasks, representing the tasks a CPU needs to do, where each letter represents a different task. Tasks could be done in any order. Each task is done in one unit of time. For each unit of time, the CPU could complete either one task or just be idle.

However, there is a non-negative integer n that represents the cooldown period between two same tasks (the same letter in the array), that is that there must be at least n units of time between any two same tasks.

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

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Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Example 2:

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Input: nums = []
Output: []

Example 3:

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Input: nums = [0]
Output: []

Constraints: