给定一个整数数组 nums 和一个目标值 target，
请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。
你可以假设每种输入只会对应一个答案。但是，你不能重复利用这个数组中同样的元素。
示例:
给定 nums = [2, 7, 11, 15], target = 9
因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1] class Solution:
def twoSum(self, nums, target):
S = set(nums)
for num in S:
pre = target - num
if num == pre:
if nums.count(num) > 1:
index1 = nums.index(num)
return [index1, nums.index(num, index1+1)]
elif pre in S:
return [nums.index(num), nums.index(target - num)]
nums = [2, 2, 7, 11, 15]
target = 9
print(Solution().twoSum(nums, target))
nums = [2, 5, 5, 11]
target = 10
print(Solution().twoSum(nums, target))
- 不能重复利用元素——转换集合进行迭代
- 依次遍历每一个元素，计算pre=tar-num
- 判断pre是否与num相等，相等且存在两个以上这样的数结束循环，返回下标
判断是否在集合中，若在结束循环，返回下标
给定一组不含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。
说明：解集不能包含重复的子集。
示例：
输入: nums = [1,2,3]
输出: [[3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], []]import itertools
class Solution:
def subsets(self, nums):
L_result = [[]]
S_c = nums.copy()
for i in range(1, len(S_c)+1):
L_result.extend(self.get_newL(i, S_c))
print(L_result)
return L_result
def get_newL(self, i, L_c):
return [i for i in itertools.combinations(L_c, i)]
a = Solution().get_newL(4, [0, 1, 2, 3, 4])
print(a)class Solution:
def subsets(self, nums):
L_result = [[]]
S_c = nums.copy()
for i in range(1, len(S_c)+1):
L_result.extend(self.get_newL(i, S_c))
print(L_result)
return L_result
def get_newL(self, i, L_c):
"""按照留的方法（就不用去了），还不错"""
L_result = [] # 存储所有结果
i_max_a = len(L_c) - i # 表示最大索引a
i_max_b = len(L_c) - 1 # 表示最大索引b
L = list(range(i)) # 临时存储结果的索引
INDEX_i = -1 # 记录滑块位置
while L[0] < i_max_a:
if L[-1] < i_max_b:
L_result.append([L_c[i] for i in L])
L[-1] += 1
else: # 到达右端
L_result.append([L_c[i] for i in L])
# 1) 进行重建
L[INDEX_i-1:] = list(range(
L[INDEX_i-1]+1,
L[INDEX_i-1]+2-INDEX_i))
# 2) INDEX_i变化
if L[-1] != i_max_b:
INDEX_i = -1
else:
INDEX_i -= 1
L_result.append([L_c[i] for i in L])
print('\n', L_result)
return L_result
a = Solution().get_newL(4, [0, 1, 2, 3, 4])
print(a)设定三个指标，控制循环发生。
给定一个 24 小时制（小时:分钟）的时间列表，找出列表中任意两个时间的最小时间差并已分钟数表示。
示例 1：
输入: ["23:59","00:00"]
输出: 1
备注:
列表中时间数在 2~20000 之间。
每个时间取值在 00:00~23:59 之间。 import datetime
class Solution:
def findMinDifference(self, timePoints):
S_timePoints = set(timePoints)
if len(S_timePoints) != len(timePoints):
return 0
L_time = [datetime.datetime.strptime(x, "%H:%M")
for x in S_timePoints]
S_time = set()
for i, t1 in enumerate(L_time):
for t2 in L_time[i+1:]:
S_time.add(self.__getSeq__(t1, t2))
return min(S_time)
def __getSeq__(self, t1, t2):
if t1 > t2:
t2_next = t2 + datetime.timedelta(days=1)
return min((t1 - t2).seconds//60,
(t2_next - t1).seconds//60)
else:
t1_next = t1 + datetime.timedelta(days=1)
return min((t2 - t1).seconds//60,
(t1_next - t2).seconds//60)
Solution().findMinDifference(["23:59", "00:00"])
Solution().findMinDifference(["00:00", "00:00", "23:58"])
Solution().findMinDifference(["00:00", "00:01", "23:58"])_
给出第一个词 first 和第二个词 second，
考虑在某些文本 text 中可能以 "first second third" 形式出现的情况，
其中 second 紧随 first 出现，third 紧随 second 出现。
对于每种这样的情况，将第三个词 "third" 添加到答案中，并返回答案。
示例 1：
输入：text = "alice is a good girl she is a good student", first = "a", second = "good"
输出：["girl","student"]
示例 2：
输入：text = "we will we will rock you", first = "we", second = "will"
输出：["we","rock"]
提示：
1 <= text.length <= 1000
text 由一些用空格分隔的单词组成，每个单词都由小写英文字母组成
1 <= first.length, second.length <= 10
first 和 second 由小写英文字母组成class Solution:
def findOcurrences(self, text, first, second):
L_result = []
L = text.split()
imax = len(L) - 1
i = 0
while i < imax - 1:
if L[i] == first and L[i+1] == second:
L_result.append(L[i+2])
i += 1
return L_result
text, first, second = ("alice is aa good girl she is a good student",
"a",
"good")
text, first, second = ("we will we will rock you",
"we",
"will")
L_result = Solution().findOcurrences(text, first, second)
print(L_result)class Solution:
def findOcurrences(self, text, first, second):
L_result = []
L_text = text.split()
text = ' ' + text
while L_text:
L_text = text.split(' %s %s ' % (first, second), 1)[1:]
# print(text, L_text)
if L_text:
for x in L_text:
L_tmp = x.split()
if L_tmp:
L_result.append(L_tmp[0])
text = ' ' + L_text[-1]
return L_result
text, first, second = ("alice is aa good girl she is a good student",
"a",
"good")
text, first, second = ("we will we will rock you",
"we",
"will")
L_result = Solution().findOcurrences(text, first, second)
print(L_result)