class Solution:
def threeSumClosest(self, nums, target):
ret = float('inf')
nums.sort()
length = len(nums)
for i in range(length - 2):
left = i + 1
right = length - 1
while left < right:
tmp = nums[i] + nums[left] + nums[right]
ret = tmp if abs(tmp - target) < abs(ret - target) else ret
if tmp == target:
return target
if tmp > target:
right -= 1
else:
left += 1
return ret
class Solution:
def threeSumClosest(self,nums,target):
ret = float('inf')
nums.sort()
length = len(nums)
for i in range(length -2):
left =i+1
right = length -1
while left<right:
tmp = nums[i] +nums[left] +nums[right]
ret = tmp if abs[tmp-target)<abs(ret-target) else ret
if tmp == target:
return target
if tmp>target:
right -=1
else:
left+=1
return ret
class Solution:
def threeSumClosest(self, nums: List[int], target: int) -> int:
nums.sort()
n = len(nums)
best = 10**7
# 根据差值的绝对值来更新答案
def update(cur):
nonlocal best
if abs(cur - target) < abs(best - target):
best = cur
# 枚举 a
for i in range(n):
# 保证和上一次枚举的元素不相等
if i > 0 and nums[i] == nums[i - 1]:
continue
# 使用双指针枚举 b 和 c
j, k = i + 1, n - 1
while j < k:
s = nums[i] + nums[j] + nums[k]
# 如果和为 target 直接返回答案
if s == target:
return target
update(s)
if s > target:
# 如果和大于 target，移动 c 对应的指针
k0 = k - 1
# 移动到下一个不相等的元素
while j < k0 and nums[k0] == nums[k]:
k0 -= 1
k = k0
else:
# 如果和小于 target，移动 b 对应的指针
j0 = j + 1
# 移动到下一个不相等的元素
while j0 < k and nums[j0] == nums[j]:
j0 += 1
j = j0
return best
