三个月算法进阶--day8

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        length = len(nums)
        if length == 3:
            return sum(nums)
        nums.sort()
        ans = []
        for i in range(length-2):
            t = i + 1
            s = length - 1
            while t < s:
                tmp = nums[i] + nums[t] + nums[s]
                if tmp < target:
                    t += 1
                elif tmp > target:
                    s -= 1
                else:
                    return target
                ans.append(tmp)
        abs_lst = [abs(target - i) for i in ans]
        return ans[abs_lst.index(min(abs_lst))]

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