375. Guess Number Higher or Lower II: minmax的一道題谋旦,一直不太會minmax的題目成榜,大概的意思就是先選擇最壞的情況羽历,然后在一堆最壞情況里選擇一個好一點的屎鳍。
class Solution(object):
def getMoneyAmount(self, n):
"""
:type n: int
:rtype: int
"""
dp = [[0 for _ in range(n+1)] for _ in range(n+1)]
# dp[i][j] is the cost to pay for guessing range i~j
for j in range(2, n+1):
for i in range(j-1, 0, -1): # j-1 ~ 1
min_value = sys.maxint
for k in range(i+1, j):
local = k + max(dp[i][k-1], dp[k+1][j])
min_value = min(local, min_value)
if j-i == 1:
min_value = i
dp[i][j] = min_value
return dp[1][n]
376. Wiggle Subsequence: 用兩個dp sequence來構(gòu)造芒粹,推導(dǎo)的時候要利用兩個dp之間的關(guān)系
377. Combination Sum IV:比較像是背包問題政钟,用target做為背包的大小闲先,只是初始化的時候屠缭,要把dp[0]初始化為1骡和,也就是說相赁,當(dāng)背包為0的時候,有一種放法慰于,就是什么都不放钮科。
378. Kth Smallest Element in a Sorted Matrix:正好是上周末九章講的題目,思路清晰得不要不要的lol
379. Design Phone Directory: 用一個deque來做婆赠,get和release就是pop和add绵脯,然后check就是loop一遍佳励。
380. Insert Delete GetRandom O(1):要點是O(1),所以用一個list來存數(shù)字蛆挫,一個hash來記錄數(shù)字的pos赃承。
382. Linked List Random Node: 神奇的Reservoir Sampling
384. Shuffle an Array: 這又是一題關(guān)于random的題目,可以用pop任意值的方法來獲取array的random permutations悴侵。
385. Mini Parser: stack的題目瞧剖,有點暈暈乎乎
class Solution(object):
def deserialize(self, s):
"""
:type s: str
:rtype: NestedInteger
"""
stack, start = [], -1
for i, c in enumerate(s):
if c == '[':
stack.append(NestedInteger()) # 對于每一個'['創(chuàng)建一個新NestedInteger,并且加入stack
elif c == ']':
# 如果stack里有多于一個NestedInteger可免,那么pop最后一個抓于,并且把它加入到上一個中
if len(stack) > 1:
t = stack.pop()
stack[-1].add(t)
elif c in "1234567890-":
if start == -1:
start = i
if i == len(s) - 1 or s[i+1] not in "-1234567890": # 如果c是最后一個值或者如果i+1不是是一個數(shù)值
if stack:
stack[-1].add(NestedInteger(int(s[start:i + 1])))
else:
stack.append(NestedInteger(int(s[start:i + 1])))
start = -1
return stack.pop()
386. Lexicographical Numbers:感覺這種題目只能靠死記了?
class Solution(object):
def lexicalOrder(self, n):
"""
:type n: int
:rtype: List[int]
"""
ans = [1]
while len(ans) < n: # 如果ans的值空間小于n
new = ans[-1] * 10 # 最新的值是ans中的最后一個值
while new > n: # 如果新值大于n
new /= 10 # 縮小新值
new += 1 # 然后遞增
while new % 10 == 0: # deal with case like 199+1=200 when we need to restart from 2.
new /= 10
ans.append(new)
return ans