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LeetCode



There are n gas stations along a circular route, where the amount of gas at the i<sup>th</sup> station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the i<sup>th</sup> station to its next (i + 1)<sup>th</sup> station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return the starting gas station’s index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique

Example 1:

Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.


Example 2:

Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation:
You can’t start at station 0 or 1, as there is not enough gas to travel to the next station.
Let’s start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can’t travel around the circuit once no matter where you start.


Constraints:

gas.length == n cost.length == n
1 <= n <= 10<sup>5</sup> 0 <= gas[i], cost[i] <= 10<sup>4</sup>

题目大意：

N/A

算法思路：

只要总gas >= 总cost，就总有一个点满足gas-cost为非负

注意事项：

  1. 先判断不合法的情况sum(gas) < sum(cost)

Python代码：

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def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
	if sum(gas) < sum(cost):
		return -1
	sum_gas, sum_cost, pos = 0, 0, 0
	for i in range(len(gas)):
		sum_gas += gas[i]
		sum_cost += cost[i]
		if sum_gas < sum_cost:
			pos = i + 1
			sum_gas = 0
			sum_cost = 0
	return pos

算法分析：

时间复杂度为O(n)，空间复杂度O(1)

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LeetCode



You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.


Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.


Constraints:

1 <= nums.length <= 100 0 <= nums[i] <= 400

算法思路：

N/A

注意事项：

  1. 循环不是模板中的1开始，而是从2开始，因为i-2>=0

Python代码：

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def rob(self, nums: List[int]) -> int:
	dp = [0] * (len(nums) + 1)
	dp[1] = nums[0]
	for i in range(2, len(dp)):
		dp[i] = max(dp[i-1], dp[i-2] + nums[i - 1])
	return dp[-1]

算法分析：

时间复杂度为O(n)，空间复杂度O(1)

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LeetCode



You are given an integer array nums. The range of a subarray of nums is the difference between the largest and smallest element in the subarray.

Return the sum of all subarray ranges of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.


Example 2:

Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.


Example 3:

Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.


Constraints:

1 <= nums.length <= 1000 -10<sup>9</sup> <= nums[i] <= 10<sup>9</sup>

题目大意：

求所有子数组的最大值最小值之差的和

Stack算法思路：

参考Leetcode 907，分别求子数组最小值的相反数，子数组的最大值，这两个值的和即为所求

注意事项：

  1. 最小值用递增栈，最大值用递减栈

Python代码：

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def subArrayRanges(self, nums: List[int]) -> int:
	arr = list(nums)
	arr.insert(0, -sys.maxsize)
	arr.append(-sys.maxsize)
	stack, res, = [], 0
	for i in range(len(arr)):
		while stack and arr[i] < arr[stack[-1]]:
			prev_idx = stack.pop()
			res -= arr[prev_idx] * (prev_idx - stack[-1]) * (i - prev_idx)
		stack.append(i)

	arr = list(nums)
	arr.insert(0, sys.maxsize)
	arr.append(sys.maxsize)

	for i in range(len(arr)):
		while stack and arr[i] > arr[stack[-1]]:
			prev_idx = stack.pop()
			res += arr[prev_idx] * (prev_idx - stack[-1]) * (i - prev_idx)
		stack.append(i)
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(n)

累计和算法II解题思路：

比暴力法稍优，两重循环覆盖所有子数组[i, j]，每轮循环得到最大最小值，然后O(1)内求该区间内所有最大最小值差值和。

Python代码：

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def subArrayRanges(self, nums: List[int]) -> int:
	sum = 0
	for i in range(len(nums) - 1):
		min_value, max_value = nums[i], nums[i]
		for j in range(i + 1, len(nums)):
			min_value = min(min_value, nums[j])
			max_value = max(max_value, nums[j])
			sum += max_value - min_value
	return sum

算法分析：

时间复杂度为O(n^2)，空间复杂度O(1)

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LeetCode



Fruits are available at some positions on an infinite x-axis. You are given a 2D integer array fruits where fruits[i] = [position<sub>i</sub>, amount<sub>i</sub>] depicts amount<sub>i</sub> fruits at the position position<sub>i</sub>. fruits is already sorted by position<sub>i</sub> in ascending order, and each position<sub>i</sub> is unique.

You are also given an integer startPos and an integer k. Initially, you are at the position startPos. From any position, you can either walk to the left or right. It takes one step to move one unit on the x-axis, and you can walk at most k steps in total. For every position you reach, you harvest all the fruits at that position, and the fruits will disappear from that position.

Return the maximum total number of fruits you can harvest.

Example 1:



Input: fruits = [[2,8],[6,3],[8,6]], startPos = 5, k = 4
Output: 9
Explanation:
The optimal way is to:
- Move right to position 6 and harvest 3 fruits
- Move right to position 8 and harvest 6 fruits
You moved 3 steps and harvested 3 + 6 = 9 fruits in total.


Example 2:



Input: fruits = [[0,9],[4,1],[5,7],[6,2],[7,4],[10,9]], startPos = 5, k = 4
Output: 14
Explanation:
You can move at most k = 4 steps, so you cannot reach position 0 nor 10.
The optimal way is to:
- Harvest the 7 fruits at the starting position 5
- Move left to position 4 and harvest 1 fruit
- Move right to position 6 and harvest 2 fruits
- Move right to position 7 and harvest 4 fruits
You moved 1 + 3 = 4 steps and harvested 7 + 1 + 2 + 4 = 14 fruits in total.


Example 3:



Input: fruits = [[0,3],[6,4],[8,5]], startPos = 3, k = 2
Output: 0
Explanation:
You can move at most k = 2 steps and cannot reach any position with fruits.


Constraints:

1 <= fruits.length <= 10<sup>5</sup> fruits[i].length == 2
`0 <= startPos, positioni <= 2 105*positioni-1 < positionifor anyi > 0(**0-indexed**) *1 <= amounti <= 104*0 <= k <= 2 * 105`

题目大意：

向左向右在规定步数内采集每一格的水果，求最大水果数

算法思路：

一开始考虑用BFS，但由于每个点可以走两次，如先往左再往右，所以不能用BFS
每个点不能走3次，因为贪婪法。所以只要计算单向路径的水果数，单向路径水果数只要计算[startPos - k - 1, startPos + k + 1]的这个区间即可
然后重复路径的范围是[0, k/2 + 1], 枚举这些值然后用presum得到单向路径水果数。

注意事项：

  1. 先判断不合法的情况sum(gas) < sum(cost)

Python代码：

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def maxTotalFruits(self, fruits: List[List[int]], startPos: int, k: int) -> int:
	pos_to_fruits = collections.defaultdict(int)
	for pair in fruits:
		pos_to_fruits[pair[0]] = pair[1]
	presum = collections.defaultdict(int)
	presum[startPos - k - 1] = pos_to_fruits[startPos - k - 1]
	for i in range(startPos - k, startPos + k + 1):
		presum[i] += presum[i-1] + pos_to_fruits[i]
	res = 0
	for i in range(k//2 + 1):
		res = max(res, presum[startPos + k - i * 2] - presum[startPos - i - 1])
		res = max(res, presum[startPos + i] - presum[startPos - k + i * 2 - 1])
	return res

算法分析：

时间复杂度为O(k + n)，空间复杂度O(n + k)

Posted on Edited on Disqus:

LeetCode



Given an integer array nums and an integer k, return the k<sup>th</sup> largest element in the array.

Note that it is the k<sup>th</sup> largest element in the sorted order, not the k<sup>th</sup> distinct element.

Example 1:

Input: nums = [3,2,1,5,6,4], k = 2
Output: 5


Example 2:

Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
Output: 4


Constraints:

1 <= k <= nums.length <= 10<sup>4</sup> -10<sup>4</sup> <= nums[i] <= 10<sup>4</sup>

题目大意：

求第k大的数(1th index)

Heap算法思路：

求第k个最大也就是用最小堆(大->小)

注意事项：

N/A

Python代码：

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def findKthLargest(self, nums: List[int], k: int) -> int:
	res = []  # min heap
	for i in range(len(nums)):
		if i < k:
			heapq.heappush(res, nums[i])
		elif nums[i] > res[0]:
			heapq.heapreplace(res, nums[i])
	return res[0]

算法分析：

时间复杂度为O(nlogk)，空间复杂度O(k)

Quickselect算法II解题思路：

N/A

注意事项：

  1. 递归调用仍用m，而不是跟pivot_pos相关，因为m是下标位置
  2. partition中range用[start, end)而不是len

Python代码：

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def findKthLargest(self, nums: List[int], k: int) -> int:
	m = len(nums) - k
	return self.quick_select(nums, 0, len(nums) - 1, m)

def quick_select(self, nums, start, end, m):
	if start > end:
		return -1
	pivot_pos = self.partition(nums, start, end)
	if m == pivot_pos:
		return nums[pivot_pos]
	elif m < pivot_pos:
		return self.quick_select(nums, start, pivot_pos - 1, m)
	else:
		return self.quick_select(nums, pivot_pos + 1, end, m)  # remember use m not related to pivot_pos

def partition(self, nums, start, end):
	pivot, no_smaller_index = nums[end], start
	for i in range(start, end):  # remember use start and end not len
		if nums[i] < pivot:
			nums[i], nums[no_smaller_index] = nums[no_smaller_index], nums[i]
			no_smaller_index += 1
	nums[no_smaller_index], nums[end] = nums[end], nums[no_smaller_index]
	return no_smaller_index

算法分析：

T(n) = T(n/2)+n, 时间复杂度为O(n)，空间复杂度O(1)

排序算法III解题思路：

先排序

算法分析：

时间复杂度为O(nlogn)，空间复杂度O(1)

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