所有題目均採用python進行完成
def
binary_search
(num, target)
:# 判斷是否為空
iflen
(num)==0
:return-1
left =
0 right =
len(num)-1
while left +
1< right:
mid =
(left + right)//2
if num[mid]
== target:
right = mid
elif num[mid]
< target:
left = mid
elif num[mid]
> target:
right = mid
# 判斷邊界條件
if num[left]
== target:
return left
if num[right]
== target:
return right
return
-1
def
search
(num):if
len(num)==0
:return-1
left =
0 right =
len(num)-1
while left +
1< right:
if num[left]
< num[right]
:return num[left]
mid = left +
(right - left)//2
if num[left]
<= num[mid]
: left = mid +
1else
: right = mid
return num[left]
if num[left]
< num[right]
else num[right]
def
search_target
(num, target):if
len(num)==0
:return-1
left, right =0,
len(num)-1
while left +
1< right:
mid = left +
(right - left)//2
if num[mid]
== target:
return mid
if num[left]
< num[mid]
:if num[left]
<= target and target <= num[mid]
: right = mid
else
: left = mid
else
:if num[mid]
<= target and target <= num[right]
: left = mid
else
: right = mid
if num[left]
== target:
return left
if num[right]
== target:
return right
return
-1
def
search_insert_position
(num, target):if
len(num)==0
:return
0 left, right =0,
len(num)-1
while left +
1< right:
mid = left +
(right - left)//2
if num[mid]
== target:
return mid
if num[mid]
> target:
right = mid
else
: left = mid
if num[left]
>= target:
return left
if num[right]
>= target:
return right
return right +
1
思路：找到第一次出現目標和最後一次出現目標的位置，寫兩個二分法
def
search_area
(num, target):if
len(num)==0
:return(-
1,-1
)# left bound
left, right =0,
len(num)-1
while left +
1< right:
mid = left +
(right - left)//2
if num[mid]
== target:
right = mid
elif num[mid]
> target:
right = mid
else
: left = mid
if num[left]
== target:
lbound = left
elif num[right]
== target:
lbound = right
else
:return(-
1,-1
)# right bound
left, right =0,
len(num)-1
while left +
1< right:
mid = left +
(right - left)//2
if num[mid]
== target:
left = mid
elif num[mid]
> target:
right = mid
else
: left = mid
if num[left]
== target:
rbound = left
elif num[right]
== target:
rbound = right
else
:return(-
1,-1
)return
(lbound, rbound)
首先確定左邊 or 右邊第乙個字元的位置
def
search_empty
(num, target):if
len(num)==0
:return-1
left, right,=0
,len
(num)-1
while left +
1< right:
while left +
1< right and left =='':
if num[left]
==''
: left +=
1if right < left:
return-1
mid = left +
(right - left)//2
if num[mid]
== target:
return mid
if num[mid]
< target:
left = mid +
1else
: right = mid -
1if num[left]
== target:
return left
if num[right]
== target:
return right
return
-1
每日程式設計題 03
所有題目實現都是採用python 在乙個m行n列的矩陣中，如果查詢到乙個元素為 0 將其所在的行和列的元素都設為 0 需要考慮乙個記錄矩陣，一般都會考慮m行n列的，但是可以行列分開考慮，這樣會減少空間複雜度，只需要 m n def myzero matrix m none len matrix n ...
每日程式設計題 06
所有題目還是採用python進行解答的哦！今天的題目相對來說比較簡單，也有很多種解決方法，我只是給出了一些方法，大家可以多多查查，希望一起交流學習。def forsolve n for迴圈 s 1if n 0 return 1else for i in range 1 n 1 s i return ...
每日程式設計題 11
題目均採用python進行完成 已知供暖裝置和房屋的位置，給出供暖裝置的最小半徑 注意 供暖裝置需要排序 可以使用python中的bisect找到需要插入的索引值 from bisect import bisect deffind radius house,heater 供暖裝置排序 heater....