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Merge pull request youngyangyang04#2078 from jianghongcheng/master
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python 二叉树 章节 持续更新。
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@youngyangyang04
youngyangyang04 authored May 7, 2023
2 parents af4b5d0 + 6e88b20 commit c80796c
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49 changes: 48 additions & 1 deletion problems/0001.两数之和.md
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Original file line number Diff line number Diff line change
Expand Up @@ -151,7 +151,7 @@ public int[] twoSum(int[] nums, int target) {
```

Python：

（版本一） 使用字典
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
Expand All @@ -163,6 +163,53 @@ class Solution:
records[value] = index # 遍历当前元素，并在map中寻找是否有匹配的key
return []
```
（版本二）使用集合
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
#创建一个集合来存储我们目前看到的数字
seen = set()
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
return [nums.index(complement), i]
seen.add(num)
```
（版本三）使用双指针
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
# 对输入列表进行排序
nums_sorted = sorted(nums)

# 使用双指针
left = 0
right = len(nums_sorted) - 1
while left < right:
current_sum = nums_sorted[left] + nums_sorted[right]
if current_sum == target:
# 如果和等于目标数，则返回两个数的下标
left_index = nums.index(nums_sorted[left])
right_index = nums.index(nums_sorted[right])
if left_index == right_index:
right_index = nums[left_index+1:].index(nums_sorted[right]) + left_index + 1
return [left_index, right_index]
elif current_sum < target:
# 如果总和小于目标，则将左侧指针向右移动
left += 1
else:
# 如果总和大于目标值，则将右指针向左移动
right -= 1
```
（版本四）暴力法
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if nums[i] + nums[j] == target:
return [i,j]
```

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95 changes: 53 additions & 42 deletions problems/0015.三数之和.md
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Original file line number Diff line number Diff line change
Expand Up @@ -298,61 +298,72 @@ class Solution {
```

Python：
（版本一） 双指针
```Python
class Solution:
def threeSum(self, nums):
ans = []
n = len(nums)
def threeSum(self, nums: List[int]) -> List[List[int]]:
result = []
nums.sort()
# 找出a + b + c = 0
# a = nums[i], b = nums[left], c = nums[right]
for i in range(n):
left = i + 1
right = n - 1
# 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
if nums[i] > 0:
break
if i >= 1 and nums[i] == nums[i - 1]: # 去重a

for i in range(len(nums)):
# 如果第一个元素已经大于0，不需要进一步检查
if nums[i] > 0:
return result

# 跳过相同的元素以避免重复
if i > 0 and nums[i] == nums[i - 1]:
continue
while left < right:
total = nums[i] + nums[left] + nums[right]
if total > 0:
right -= 1
elif total < 0:

left = i + 1
right = len(nums) - 1

while right > left:
sum_ = nums[i] + nums[left] + nums[right]

if sum_ < 0:
left += 1
elif sum_ > 0:
right -= 1
else:
ans.append([nums[i], nums[left], nums[right]])
# 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
while left != right and nums[left] == nums[left + 1]: left += 1
while left != right and nums[right] == nums[right - 1]: right -= 1
left += 1
result.append([nums[i], nums[left], nums[right]])

# 跳过相同的元素以避免重复
while right > left and nums[right] == nums[right - 1]:
right -= 1
while right > left and nums[left] == nums[left + 1]:
left += 1

right -= 1
return ans
left += 1

return result
```
Python (v3):
（版本二） 使用字典

```python
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
if len(nums) < 3: return []
nums, res = sorted(nums), []
for i in range(len(nums) - 2):
cur, l, r = nums[i], i + 1, len(nums) - 1
if res != [] and res[-1][0] == cur: continue # Drop duplicates for the first time.

while l < r:
if cur + nums[l] + nums[r] == 0:
res.append([cur, nums[l], nums[r]])
# Drop duplicates for the second time in interation of l & r. Only used when target situation occurs, because that is the reason for dropping duplicates.
while l < r - 1 and nums[l] == nums[l + 1]:
l += 1
while r > l + 1 and nums[r] == nums[r - 1]:
r -= 1
if cur + nums[l] + nums[r] > 0:
r -= 1
result = []
nums.sort()
# 找出a + b + c = 0
# a = nums[i], b = nums[j], c = -(a + b)
for i in range(len(nums)):
# 排序之后如果第一个元素已经大于零，那么不可能凑成三元组
if nums[i] > 0:
break
if i > 0 and nums[i] == nums[i - 1]: #三元组元素a去重
continue
d = {}
for j in range(i + 1, len(nums)):
if j > i + 2 and nums[j] == nums[j-1] == nums[j-2]: # 三元组元素b去重
continue
c = 0 - (nums[i] + nums[j])
if c in d:
result.append([nums[i], nums[j], c])
d.pop(c) # 三元组元素c去重
else:
l += 1
return res
d[nums[j]] = j
return result
```

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84 changes: 41 additions & 43 deletions problems/0018.四数之和.md
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Original file line number Diff line number Diff line change
Expand Up @@ -207,72 +207,70 @@ class Solution {
```

Python：
(版本一) 双指针
```python
# 双指针法
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:

nums.sort()
n = len(nums)
res = []
for i in range(n):
if i > 0 and nums[i] == nums[i - 1]: continue # 对nums[i]去重
for k in range(i+1, n):
if k > i + 1 and nums[k] == nums[k-1]: continue # 对nums[k]去重
p = k + 1
q = n - 1

while p < q:
if nums[i] + nums[k] + nums[p] + nums[q] > target: q -= 1
elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1
result = []
for i in range(n):
if nums[i] > target and nums[i] > 0 and target > 0:# 剪枝（可省）
break
if i > 0 and nums[i] == nums[i-1]:# 去重
continue
for j in range(i+1, n):
if nums[i] + nums[j] > target and target > 0: #剪枝（可省）
break
if j > i+1 and nums[j] == nums[j-1]: # 去重
continue
left, right = j+1, n-1
while left < right:
s = nums[i] + nums[j] + nums[left] + nums[right]
if s == target:
result.append([nums[i], nums[j], nums[left], nums[right]])
while left < right and nums[left] == nums[left+1]:
left += 1
while left < right and nums[right] == nums[right-1]:
right -= 1
left += 1
right -= 1
elif s < target:
left += 1
else:
res.append([nums[i], nums[k], nums[p], nums[q]])
# 对nums[p]和nums[q]去重
while p < q and nums[p] == nums[p + 1]: p += 1
while p < q and nums[q] == nums[q - 1]: q -= 1
p += 1
q -= 1
return res
right -= 1
return result

```
(版本二) 使用字典

```python
# 哈希表法
class Solution(object):
def fourSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[List[int]]
"""
# use a dict to store value:showtimes
hashmap = dict()
for n in nums:
if n in hashmap:
hashmap[n] += 1
else:
hashmap[n] = 1
# 创建一个字典来存储输入列表中每个数字的频率
freq = {}
for num in nums:
freq[num] = freq.get(num, 0) + 1

# good thing about using python is you can use set to drop duplicates.
# 创建一个集合来存储最终答案，并遍历4个数字的所有唯一组合
ans = set()
# ans = [] # save results by list()
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
for k in range(j + 1, len(nums)):
val = target - (nums[i] + nums[j] + nums[k])
if val in hashmap:
# make sure no duplicates.
if val in freq:
# 确保没有重复
count = (nums[i] == val) + (nums[j] == val) + (nums[k] == val)
if hashmap[val] > count:
ans_tmp = tuple(sorted([nums[i], nums[j], nums[k], val]))
ans.add(ans_tmp)
# Avoiding duplication in list manner but it cause time complexity increases
# if ans_tmp not in ans:
# ans.append(ans_tmp)
else:
continue
return list(ans)
# if used list() to save results, just
# return ans
if freq[val] > count:
ans.add(tuple(sorted([nums[i], nums[j], nums[k], val])))

return [list(x) for x in ans]

```

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