015 3sum
15. 3Sum
题目: https://leetcode.com/problems/3sum/
难度:
Medium
第一想法，先把nums排序，用三个loop，无法AC
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
n = len(nums)
res = []
nums.sort()
for i in range(n):
for j in range(i,n):
for k in range(j,n):
if nums[i] + nums[j] + nums[k] == 0 and j != i and k != j and k != i:
curRes = [nums[i],nums[j],nums[k]]
if curRes not in res:
res.append(curRes)
return res
然后查了一下2sum，用2sum的花样，因为要排除重复以及输出是按照从小到大的输出:但是还是超时
class Solution(object): # 此法也超时
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
def twoSum(nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
lookup = {}
for num in nums:
if target - num in lookup:
if (-target ,target - num, num) not in res:
res.append((-target ,target - num, num))
lookup[num] = target - num
n = len(nums)
nums.sort()
res = []
for i in range(n):
twoSum(nums[i+1:], 0-nums[i])
return [list(i) for i in res]
谷歌看别人的代码，思路非常清晰的,运行起来比直接调用 Two Sum快.
清晰的思路：
- 排序
- 固定左边，如果左边重复，继续
- 左右弄边界，去重，针对不同的左右边界情况处理
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
n, res = len(nums), []
nums.sort()
for i in range(n):
if i > 0 and nums[i] == nums[i-1]: # 因为i=0这个元素会直接往下执行
continue
l, r = i+1, n-1
while l < r:
tmp = nums[i] + nums[l] + nums[r]
if tmp == 0:
res.append([nums[i], nums[l], nums[r]])
l += 1
r -= 1
while l < r and nums[l] == nums[l-1]:
l += 1
while l < r and nums[r] == nums[r+1]:
r -= 1
elif tmp > 0:
r -= 1
else:
l += 1
return res
