lintcode亚麻九题
628.maximum-subtree
Given a binary tree, find the subtree with maximum sum. Return the root of the subtree.
思路很简单，分治就可以
import sys
class Solution:
# @param {TreeNode} root the root of binary tree
# @return {int} the maximum weight node
def findSubtree(self, root):
# Write your code here
self.max_value = -sys.maxint
self.max_root = None
self.helper(root)
return self.max_root
def helper(self, root):
if not root:
return 0
left = self.helper(root.left)
right = self.helper(root.right)
result = root.val + left + right
if result > self.max_value:
self.max_value = result
self.max_root = root
return result
627.Longest Palindrome
Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
大小写敏感。注意是构建，所以统计出每个字符出现次数，然后对偶数求和，如果存在奇数，-1求和，最后+1
class Solution:
# @param {string} s a string which consists of lowercase or uppercase letters
# @return {int} the length of the longest palindromes that can be built
def longestPalindrome(self, s):
# Write your code here
cache = {}
result = 0
for word in s:
if word in cache:
cache[word] += 1
else:
cache[word] = 1
flag = False
for key in cache:
if cache[key] % 2 == 0:
result += cache[key]
else:
result += cache[key] - 1
flag = True
if flag:
result += 1
return result
626.Rectangle Overlap
Given two rectangles, find if the given two rectangles overlap or not.
给出的是两个矩形的左上角和右下角，取两个矩形端点比较靠向中心的（交集），判断这个交集是否为空
# Definition for a point.
# class Point:
# def __init__(self, a=0, b=0):
# self.x = a
# self.y = b
class Solution:
# @param {Point} l1 top-left coordinate of first rectangle
# @param {Point} r1 bottom-right coordinate of first rectangle
# @param {Point} l2 top-left coordinate of second rectangle
# @param {Point} r2 bottom-right coordinate of second rectangle
# @return {boolean} true if they are overlap or false
def doOverlap(self, l1, r1, l2, r2):
# Write your code here
p1 = Point(max(l1.x, l2.x), min(l1.y, l2.y))
p2 = Point(min(r1.x, r2.x), max(r1.y, r2.y))
if p1.x <= p2.x and p1.y >= p2.y:
return True
return False
判断重叠很麻烦，但判断不重叠很简单的：
# Definition for a point.
# class Point:
# def __init__(self, a=0, b=0):
# self.x = a
# self.y = b
class Solution:
# @param {Point} l1 top-left coordinate of first rectangle
# @param {Point} r1 bottom-right coordinate of first rectangle
# @param {Point} l2 top-left coordinate of second rectangle
# @param {Point} r2 bottom-right coordinate of second rectangle
# @return {boolean} true if they are overlap or false
def doOverlap(self, l1, r1, l2, r2):
# Write your code here
if l1.x > r2.x or l2.x > r1.x:
return False
if r1.y > l2.y or r2.y > l1.y:
return False
return True
604.window sum
一个滑动窗遍历数组，求滑动窗的和。
减掉移出的，加入移入的。
class Solution:
# @param nums {int[]} a list of integers
# @param k {int} size of window
# @return {int[]} the sum of element inside the window at each moving
def winSum(self, nums, k):
# Write your code here
result = []
if not nums:
return result
temp = None
for i in range(len(nums) - k + 1):
if not result:
result.append(sum(nums[i : i + k]))
else:
result.append(result[-1] - nums[i - 1] + nums[i + k - 1])
return result
616.course-schedule-ii
先修课程，典型的拓扑排序
class Solution:
# @param {int} numCourses a total of n courses
# @param {int[][]} prerequisites a list of prerequisite pairs
# @return {int[]} the course order
def findOrder(self, numCourses, prerequisites):
# Write your code here
nodes = {}
for i in range(numCourses):
nodes[i] = {
"pre" : 0,
"after" : []
}
for p in prerequisites:
nodes[p[1]]["after"].append(p[0])
nodes[p[0]]["pre"] += 1
result = []
queue = []
for key in nodes:
if nodes[key]["pre"] == 0:
queue.append(key)
while queue:
key = queue.pop(0)
result.append(key)
for after in nodes[key]["after"]:
nodes[after]["pre"] -= 1
if nodes[after]["pre"] == 0:
queue.append(after)
if len(result) == numCourses:
return result
return []
613.High Five
每个学生有两个属性 id 和 scores。找到每个学生最高的5个分数的平均值。
'''
Definition for a Record
class Record:
def __init__(self, id, score):
self.id = id
self.score = score
'''
import heapq
class Solution:
# @param {Record[]} results a list of <student_id, score>
# @return {dict(id, average)} find the average of 5 highest scores for each person
# <key, value> (student_id, average_score)
def highFive(self, results):
# Write your code here
score = {}
for r in results:
if r.id in score:
score[r.id].append(r.score)
else:
score[r.id] = [r.score]
answer = {}
for id in score:
answer[id] = sum(heapq.nlargest(5, score[id])) / 5.0
return answer
612.k closest points
找到离目标点最近的k个点。heapq + functools.cmp_to_key
# Definition for a point.
# class Point:
# def __init__(self, a=0, b=0):
# self.x = a
# self.y = b
import heapq
from functools import cmp_to_key
class Solution:
# @param {Pint[]} points a list of points
# @param {Point} origin a point
# @param {int} k an integer
# @return {Pint[]} the k closest points
def kClosest(self, points, origin, k):
# Write your code here
def cmp(x1, x2):
distance = ((x1.x - origin.x) ** 2 + (x1.y - origin.y) ** 2) \
- ((x2.x - origin.x) ** 2 + (x2.y - origin.y) ** 2)
if distance < 0:
return -1
elif distance == 0:
if x1.x < x2.x:
return -1
elif x1.x == x2.x:
if x1.y < x2.y:
return -1
elif x1.y == x2.y:
return 0
else:
return 1
else:
return 1
else:
return 1
return heapq.nsmallest(k, points, key = cmp_to_key(cmp))
105.Copy List with Random Pointer
带有随机指针的链表，做一次完全的拷贝。
下次尝试O(1)空间复杂度
# Definition for singly-linked list with a random pointer.
# class RandomListNode:
# def __init__(self, x):
# self.label = x
# self.next = None
# self.random = None
class Solution:
# @param head: A RandomListNode
# @return: A RandomListNode
def copyRandomList(self, head):
# write your code here
nodes = {}
# copy nodes
cur = head
while cur:
nodes[cur.label] = RandomListNode(cur.label)
cur = cur.next
# copy next
cur = head
while cur.next:
nodes[cur.label].next = nodes[cur.next.label]
cur = cur.next
# copy random
cur = head
while cur:
if cur.random:
nodes[cur.label].random = nodes[cur.random.label]
cur = cur.next
return nodes[head.label]
629.minimum spanning tree
最小生成树，Prim算法思想类似于BFS
'''
Definition for a Connection
class Connection:
def __init__(self, city1, city2, cost):
self.city1, self.city2, self.cost = city1, city2, cost
'''
import heapq
class Solution:
# @param {Connection[]} connections given a list of connections
# include two cities and cost
# @return {Connection[]} a list of connections from results
def lowestCost(self, connections):
# Write your code here
heap = []
result = []
cities = {}
cons = {}
for c in connections:
cons[(c.cost, c.city1, c.city2)] = c
cities.setdefault(c.city1, []).append((c.cost, c.city1, c.city2))
cities.setdefault(c.city2, []).append((c.cost, c.city1, c.city2))
visited = set([connections[0].city1])
heap = [c for c in cities[connections[0].city1]]
heapq.heapify(heap)
while heap:
c = heapq.heappop(heap)
while c[1] in visited and c[2] in visited and heap:
c = heapq.heappop(heap)
if c[1] in visited and c[2] in visited:
break
result.append(c)
new_city = c[1] if c[1] not in visited else c[2]
visited.add(new_city)
for new_c in cities[new_city]:
if new_c[1] in visited and new_c[2] in visited:
continue
heapq.heappush(heap, new_c)
result.sort()
return [cons[r] for r in result] if len(visited) == len(cities) else []
