【LeetCode】【16.3sum closest】（python版）

最新推荐文章于 2020-01-15 11:52:22 发布

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最新推荐文章于 2020-01-15 11:52:22 发布

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分类专栏：数组 LeetCode 文章标签：面试算法

本文链接：https://blog.youkuaiyun.com/qq_20141867/article/details/80946259

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本文介绍了一个算法问题——寻找数组中三个整数的组合，使其和最接近给定的目标值。通过排序和双指针技巧，有效地解决了这个问题，并提供了Python实现。

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Description：
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

思路：
本题要求的是找到三个数，满足三数之和与target的距离最近，思路与3sum大体相同，仍然是固定一个数，用两指针双向往中间走，记录并维护当前三数之和与target的最小距离的差值mindiff，实现代码如下：

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        mindiff= float("inf")
        n = len(nums)
        for i in range(n - 2):
            if i > 0 and nums[i] == nums[i - 1]: continue
            low = i + 1
            high = n - 1
            while low < high:
                cursum = nums[low] + nums[high] + nums[i]
                if cursum == target:
                    while low < high and nums[low] == nums[low + 1]:
                        low = low + 1
                    while low < high and nums[high] == nums[high - 1]:
                        high = high - 1
                    high = high - 1
                    low = low + 1
                elif cursum > target:
                    high = high - 1
                else:
                    low = low + 1
                # 比较mindiff和当前计算差值的绝对值
                if abs(mindiff) > abs(target - cursum):
                    # 每次记录的都是target - 当前三数之和
                    mindiff = target - cursum
        return target - mindiff