leetcode刷题-16-最接近的三数之和

问题描述

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。

示例

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).

实现

双指针两侧扩展

双指针从目标值开始，同时向两侧扩展，取最先遇到的三个数相加

def three_sum_clostest(nums, target):
    """
    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。
    """
    result_sum = 0
    pre_distance = {}
    if target<0:
        max_width = len(nums) - target
    else:
        max_width = len(nums) + target
    current_nums_num = 0
    break_flag = False
    for x in nums:
        # print(x)
        current_dis = (target - x)

        if current_dis < 0:
            current_dis = - current_dis

        if pre_distance.__contains__(x) and current_dis != 0:
            pre_distance[x] += current_dis
        elif pre_distance.__contains__(x) and current_dis == 0:
            pre_distance[x] += 1
        elif current_dis != 0:
            pre_distance[x] = current_dis
        elif current_dis == 0:
            pre_distance[x] = 1

    print(pre_distance)

    if pre_distance.__contains__(target):
        if pre_distance[target] >= 3:
            result_sum = 3 * target
            break_flag = True
        elif pre_distance[target] < 3:
            result_sum = pre_distance[target] * target
            current_nums_num = pre_distance[target]

    for i in range(1, max_width):
        if break_flag:
            break
        if current_nums_num == 3:
            break
        if pre_distance.__contains__((target-i)):
            distance = pre_distance[(target-i)]
            if distance == i:
                result_sum += (target-i)
                current_nums_num += 1
            elif (distance/i) >= (3-current_nums_num):
                result_sum += (3-current_nums_num) * (target-i)
                break
            elif (distance/i) < (3-current_nums_num):
                for j in range(1, (3-current_nums_num)):
                    if current_nums_num == 3:
                        break
                    result_sum += (target-i)
                    current_nums_num += 1

        if pre_distance.__contains__((target+i)):
            distance = pre_distance[(target+i)]
            print(distance)
            if distance == i:
                result_sum += (target+i)
                current_nums_num += 1
            elif (distance/i) >= (3-current_nums_num):
                result_sum += (3-current_nums_num) * ((target+i))
                break
            elif (distance/i) < (3-current_nums_num):
                for j in range(1, (3-current_nums_num)):
                    if current_nums_num == 3:
                        break
                    result_sum += (target+i)
                    current_nums_num += 1

    return result_sum