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16. 最接近的三数之和

English Version

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。

 

示例：

输入：nums = [-1,2,1,-4], target = 1
输出：2
解释：与 target 最接近的和是 2 (-1 + 2 + 1 = 2) 。

 

提示：

• 3 <= nums.length <= 10^3
• -10^3 <= nums[i] <= 10^3
• -10^4 <= target <= 10^4

解法

双指针解决。

Python3

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        def twoSumClosest(nums, start, end, target):
            res = 0
            diff = 10000
            while start < end:
                val = nums[start] + nums[end]
                if val == target:
                    return val
                if abs(val - target) < diff:
                    res = val
                    diff = abs(val - target)
                if val < target:
                    start += 1
                else:
                    end -= 1
            return res

        nums.sort()
        res, n = 0, len(nums)
        diff = 10000
        for i in range(n - 2):
            t = twoSumClosest(nums, i + 1, n - 1, target - nums[i])
            if abs(nums[i] + t - target) < diff:
                res = nums[i] + t
                diff = abs(nums[i] + t - target)
        return res

Java

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int res = 0;
        int n = nums.length;
        int diff = Integer.MAX_VALUE;
        for (int i = 0; i < n - 2; ++i) {
            int t = twoSumClosest(nums, i + 1, n - 1, target - nums[i]);
            if (Math.abs(nums[i] + t - target) < diff) {
                res = nums[i] + t;
                diff = Math.abs(nums[i] + t - target);
            }
        }
        return res;
    }

    private int twoSumClosest(int[] nums, int start, int end, int target) {
        int res = 0;
        int diff = Integer.MAX_VALUE;
        while (start < end) {
            int val = nums[start] + nums[end];
            if (val == target) {
                return val;
            }
            if (Math.abs(val - target) < diff) {
                res = val;
                diff = Math.abs(val - target);
            }
            if (val < target) {
                ++start;
            } else {
                --end;
            }
        }
        return res;
    }
}

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