LeetCode: 3Sum Closest

最新推荐文章于 2025-08-14 22:07:01 发布

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本文介绍了一种寻找数组中三个整数使它们的和最接近给定目标值的有效算法。通过排序和双指针技巧，将时间复杂度从O(n^3)降低到O(n^2)，实现了高效求解。

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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).

本题与之前不同，要求找的不是刚好相加等于target的三个数，而是最接近target的三个数。首先尝试暴力破解法：

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        length = len(nums)
        sum = nums[0] + nums[1] + nums[2]
        diff = abs(sum - target)
        for i in range(0, length-2):
            for j in range(i+1, length-1):
                for k in range(j+1, length):
                    temp_sum = nums[i] + nums[j] + nums[k]
                    temp_diff = abs(temp_sum - target)
                    if temp_diff < diff:
                        diff = temp_diff
                        sum = temp_sum
        return sum

果然不行，运行超时，O(n³)的复杂度无法通过。

于是尝试O(n²)复杂度的解法：

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        length = len(nums)
        sum = nums[0] + nums[1] + nums[2]
        diff = abs(sum - target)
        for i in range(length-2):
            j = i + 1
            k = length - 1
            i_sum = nums[i] + nums[j] + nums[k]
            i_diff = abs(i_sum - target)
            temp_sum = i_sum
            temp_diff = i_diff
            while j < k:
                if j == k - 1:
                    break
                if temp_sum < target:
                    j += 1
                elif temp_sum > target:
                    k -= 1
                else:
                    return target
                
                temp_sum = nums[i] + nums[j] + nums[k]
                temp_diff = abs(temp_sum - target)

                if temp_diff < i_diff:
                    i_sum = temp_sum
                    i_diff = temp_diff
                """ else:
                    break """
            if i_diff < diff:
                diff = i_diff
                sum = i_sum
        return sum

搞定。