1. 两数之和
"""
双指针，题目需要返回下标，所以记录一个数字对应的下标
"""
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
r = [[ind, num] for ind, num in enumerate(nums)]
r = sorted(r, key=lambda x: x[1])
head = 0
tail = len(r) - 1
while head < tail:
s = r[head][1] + r[tail][1]
if s == target:
return [r[head][0], r[tail][0]]
elif s > target:
tail -= 1
# 跳过重复值，去重的规则都一样（下面的也是这样）
while head < tail and r[tail][1] == r[tail+1][1]:
tail -= 1
else:
head += 1
while head < tail and r[head][1] == r[head-1][1]:
head += 1
15. 三数之和
"""
双指针，所以先固定一个数字，用双指针来找到另外两个数字。注意记得剪枝，否则一定会超时的。（被超时操控的恐惧...
"""
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
vis = set()
res = []
for i in range(len(nums)):
if nums[i] in vis:
continue
if nums[i] > 0:
break
vis.add(nums[i])
head = i + 1
tail = len(nums) - 1
while head < tail and head < len(nums) and tail < len(nums):
s = nums[i] + nums[head] + nums[tail]
if not s:
res.append([nums[i], nums[head], nums[tail]])
head += 1
tail -= 1
# 跳过重复元素
while head < tail and nums[head] == nums[head - 1]:
head += 1
while head < tail and nums[tail] == nums[tail + 1]:
tail -= 1
if s > 0:
tail -= 1
while head < tail and nums[tail] == nums[tail + 1]:
tail -= 1
if s < 0:
head += 1
while head < tail and nums[head] == nums[head - 1]:
head += 1
return res
18. 四数之和
"""
固定两个，双指针找另外两个
"""
class Solution:
def fourSum(self, nums, target):
nums.sort()
res = []
for i in range(len(nums)):
for j in range(i+1, len(nums)):
head = j + 1
tail = len(nums) - 1
while head < tail:
s = nums[i] + nums[j] + nums[head] + nums[tail]
if s == target:
res.append([nums[i], nums[j], nums[head], nums[tail]])
head += 1
tail -= 1
while head < tail and nums[head] == nums[head - 1]:
head += 1
while head < tail and nums[tail] == nums[tail + 1]:
tail -= 1
elif s > target:
tail -= 1
while head < tail and nums[tail] == nums[tail + 1]:
tail -= 1
else:
head += 1
while head < tail and nums[head] == nums[head - 1]:
head += 1
return list(set([tuple(i) for i in res]))
总结
可以看到，无论是2、3or4，都是固定除了双指针之外的元素，再用双指针去找剩下的元素，代码几乎没有改变，切记要记得剪枝。