README.md

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;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var threeSumClosest = function (nums, target) {
    let len = nums.length;
    nums.sort((a, b) => a - b);
    let diff = Infinity;
    let res;
    for (let i = 0; i < len - 2; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;
        let left = i + 1, right = len - 1;
        let cur = nums[i] + nums[i + 1] + nums[i + 2];
        if (cur > target) {
            let newDiff = Math.abs((cur - target))
            if (newDiff < diff) {
                diff = newDiff;
                res = cur;
            }
            break;
        }
        while (left < right) {
            cur = nums[i] + nums[left] + nums[right];
            if (cur === target) return target;
            let newDiff = Math.abs((cur - target))
            if (newDiff < diff) {
                diff = newDiff;
                res = cur;
            }
            if (cur < target) {
                while (nums[left] === nums[left + 1]) left++;
                left++;
                continue;
            } else {
                while (nums[right] === nums[right - 1]) right--;
                right--;
                continue;
            }
        }
    }
    return res;
};

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