LeetCode:16. 最接近的三数之和

1、题目描述

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。

  例如,给定数组 nums = [-1,2,1,-4], 和 target = 1.

  与 target 最接近的三个数的和为 2. (-1 + 2 + 1 = 2).

2、题解

2.1、解法一

class Solution:

    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        n = len(nums)
        print(n)
        if n < 3:
            return 0
        new_nums = sorted(nums)
        avg = target/3
        max_num = max(new_nums)
        min_num = min(new_nums)

        if avg >= max_num:
            return new_nums[n-3] + new_nums[n-2] + new_nums[n-1]
        elif avg <= min_num:
            return new_nums[0] + new_nums[1] + new_nums[2]
        else:
            sum_list = []
            for i in range(n-2):
                left = i+1
                right = n-1
                while left < right:
                    s = new_nums[i] + new_nums[left] + new_nums[right]
                    sum_list.append(s)
                    if new_nums[i] + new_nums[left] + new_nums[right] > target:
                        right -= 1
                    elif new_nums[i] + new_nums[left] + new_nums[right] < target:
                        left += 1
                    else:
                        return target

                    sum_list.append(s)

            return min(sum_list, key=lambda x: abs(target - x))

2.2、解法二

class Solution:

    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        n = len(nums)
        print(n)
        if n < 3:
            return 0
        nums.sort()
        print(nums)
        sum_list = []

        for i,num in enumerate(nums[0:-2]):
            l, r = i + 1, n - 1
            # 最大的情况小于target,其他的无需比较
            if num + nums[r-1] + nums[r] <= target:
                sum_list.append(num + nums[r-1] + nums[r])
            # 最小的情况大于target,其他的无需比较
            elif num + nums[l] + nums[l+1] >= target:
                sum_list.append(num + nums[l] + nums[l+1])
            # 中间情况
            else:
                while l < r:
                    print(i,l,r,num + nums[l] + nums[r])
                    sum_list.append(num + nums[l] + nums[r])
                    if num + nums[l] + nums[r] > target:
                        r -= 1
                    elif num + nums[l] + nums[r] < target:
                        l += 1
                    else:
                        return target

        print(sum_list)
        # return min(sum_list, key=lambda x: abs(target - x))
        sum_list.sort(key=lambda x: abs(target - x))
        print(sum_list)
        return sum_list[0]

  

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转载自www.cnblogs.com/bad-robot/p/10064870.html