题目: 给定一个包括 n 个整数的数组 nums 和 一个目标值 target。找出 nums 中的三个整数，使得它们的和与 target 最接近。返回这三个数的和。假定每组输入只存在唯一答案

离谱

class Solution:
    def threeSumClosest(self, nums: list[int], target: int) -> int:
        nums.sort()
        n = len(nums)
        flag = abs(nums[0] + nums[1] + nums[n-1] - target) # flag的初始值，flag用来衡量与target之间的距离
        sum = nums[0] + nums[1] + nums[n-1]  # sum的初始值
        #初始时：a, b, c = 0, 1, n-1
        for a in range(n):
            if nums[a] == nums[a-1]: #小优化，找下一个不同的值
                continue
            for b in range(a+1,n): #b从a的右边开始找
                for c in range(n-1,b,-1): #c从最右边开始逆序寻找
                    if c <= b or b <= a:
                        break
                    else:
                        flag = min(abs(flag), abs(nums[a] + nums[b] + nums[c] - target))
                        if flag == abs(nums[a] + nums[b] + nums[c] - target):
                            sum = nums[a] + nums[b] + nums[c]
                    if nums[a] + nums[b] + nums[c] - target == 0:  #最接近target的sum就是差距为0
                        sum = target
                        return sum
                    elif nums[a] + nums[b] + nums[c] - target > 0:
                        c -= 1
                        if nums[c] == nums[c + 1]:
                            continue
                    else:
                        b += 1
                        if nums[b] == nums[b - 1]:
                            continue

        return sum



if __name__ == '__main__':
    nums = [1,1,1,0]
    target = 100
    print(Solution.threeSumClosest(Solution,nums,target))

看到一个老哥的解法，我觉得很不错，分享一下

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        nums = sorted(nums)
        res = sum(nums[0:3])
        for i in range(len(nums)-1):
            l, r = i+1, len(nums) - 1
            while l < r:
                total = nums[i] + nums[l] + nums[r]
                if total < target:
                    l += 1
                elif total > target:
                    r -= 1
                elif total == target:
                    return total

                if abs(res-target) > abs(total-target):
                    res = total
        return res