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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

第二刷: f(n) = max{f(n-2)+nums[n-1], f(n-3)+nums[n-1]}, f(n)是以nums[n-1]为结尾必偷的房子，可以选择偷隔一个或隔两个。

Python代码：

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def rob(self, nums: List[int]) -> int:
	if len(nums) <= 2:
		return max(nums)
	n = len(nums)
	dp = [0] * (n + 1)
	dp[1], dp[2] = nums[0], nums[1]
	for i in range(3, len(dp)):
		dp[i] = max((dp[i-2]), dp[i-3]) + nums[i-1]
	return max(dp[n], dp[n-1])

初刷: f(n)是以nums[n-1]为结尾不必偷的房子，可以选择偷隔一个或隔两个

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



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)

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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



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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LeetCode



There is a new alien language that uses the English alphabet. However, the order among the letters is unknown to you.

You are given a list of strings words from the alien language’s dictionary, where the strings in words are sorted lexicographically by the rules of this new language.

Return a string of the unique letters in the new alien language sorted in lexicographically increasing order by the new language’s rules. If there is no solution, return "". If there are multiple solutions, return any of them.

A string s is lexicographically smaller than a string t if at the first letter where they differ, the letter in s comes before the letter in t in the alien language. If the first min(s.length, t.length) letters are the same, then s is smaller if and only if s.length < t.length.

Example 1:

Input: words = [“wrt”,”wrf”,”er”,”ett”,”rftt”]
Output: “wertf”


Example 2:

Input: words = [“z”,”x”]
Output: “zx”


Example 3:

Input: words = [“z”,”x”,”z”]
Output: “”
Explanation: The order is invalid, so return "".


Constraints:

1 <= words.length <= 100 1 <= words[i].length <= 100
* words[i] consists of only lowercase English letters.

算法思路：

N/A

注意事项：

  1. 题目要求： 空字符顺序前于非空字母，否则字典不合法，如abc -> ab, c不能前于空字符，无解。简而言之，后面单词不能是前面的前缀return False if len(word) > len(word2) else True
  2. 模板问题： graph要含所有节点，包括没有边的节点。否则结果会有遗漏graph = Counter({c: [] for word in words for c in word})
  3. 模板问题： in_degree初始化要对所有节点赋0， in_degree[c] = 0。in_degree = collections.defaultdict(int)并不能产生key
  4. 模板问题： 第四步判断是否含循环必不可少，题目要求可能不合法，return res if len(graph) == len(res) else ‘’
  5. 语法错误： graph.items()记得加items。res是str不是list

Python代码：

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def alienOrder(self, words: List[str]) -> str:
	# graph = collections.defaultdict(list)
	graph = Counter({c: [] for word in words for c in word})
	for i in range(1, len(words)):
		if not self.populate_one_order(words[i - 1], words[i], graph):
			return ''
	in_degree = collections.defaultdict(int)
	for c in graph.keys():
		in_degree[c] = 0
	for key, li in graph.items():
		for j in range(len(li)):
			in_degree[li[j]] += 1
	res = ''
	queue = deque([node for node, in_degree_num in in_degree.items() if in_degree_num == 0])
	while queue:
		node = queue.popleft()
		res += node
		for neighbor in graph[node]:
			in_degree[neighbor] -= 1
			if in_degree[neighbor] == 0:
				queue.append(neighbor)
	return res if len(graph) == len(res) else ''

def populate_one_order(self, word, word2, graph):
	for j in range(min(len(word), len(word2))):
		if word[j] != word2[j]:
			graph[word[j]].append(word2[j])
			return True
	return False if len(word) > len(word2) else True

算法分析：

时间复杂度为O(V + E)，空间复杂度O(O + E)，V为节点数，E为边数，n为单词数，L为最长单词长度，O = nL， E = L，而空间复杂度图的空间为O + E，而queue空间最多为26条边，26个字母，in_degree空间为O(V)。所以总的时间复杂度为O(nL)，空间复杂度O(nL)

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