README_EN.md

16. 3Sum Closest

中文文档

Description

Given an integer array nums of length n 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 1:

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Example 2:

Input: nums = [0,0,0], target = 1
Output: 0
Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0).

 

Constraints:

• 3 <= nums.length <= 500
• -1000 <= nums[i] <= 1000
• -104 <= target <= 104

Solutions

Python3

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        nums.sort()
        n = len(nums)
        ans = inf
        for i, v in enumerate(nums):
            j, k = i + 1, n - 1
            while j < k:
                t = v + nums[j] + nums[k]
                if t == target:
                    return t
                if abs(t - target) < abs(ans - target):
                    ans = t
                if t > target:
                    k -= 1
                else:
                    j += 1
        return ans

Java

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int ans = 1 << 30;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            int j = i + 1, k = n - 1;
            while (j < k) {
                int t = nums[i] + nums[j] + nums[k];
                if (t == target) {
                    return t;
                }
                if (Math.abs(t - target) < Math.abs(ans - target)) {
                    ans = t;
                }
                if (t > target) {
                    --k;
                } else {
                    ++j;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int ans = 1 << 30;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int j = i + 1, k = n - 1;
            while (j < k) {
                int t = nums[i] + nums[j] + nums[k];
                if (t == target) return t;
                if (abs(t - target) < abs(ans - target)) ans = t;
                if (t > target)
                    --k;
                else
                    ++j;
            }
        }
        return ans;
    }
};

Go

func threeSumClosest(nums []int, target int) int {
	sort.Ints(nums)
	ans := 1 << 30
	n := len(nums)
	for i, v := range nums {
		j, k := i+1, n-1
		for j < k {
			t := v + nums[j] + nums[k]
			if t == target {
				return t
			}
			if abs(t-target) < abs(ans-target) {
				ans = t
			}
			if t > target {
				k--
			} else {
				j++
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var threeSumClosest = function (nums, target) {
    nums.sort((a, b) => a - b);
    let ans = 1 << 30;
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        let j = i + 1;
        let k = n - 1;
        while (j < k) {
            const t = nums[i] + nums[j] + nums[k];
            if (t === target) {
                return t;
            }
            if (Math.abs(t - target) < Math.abs(ans - target)) {
                ans = t;
            }
            if (t > target) {
                --k;
            } else {
                ++j;
            }
        }
    }
    return ans;
};

TypeScript

function threeSumClosest(nums: number[], target: number): number {
    nums.sort((a, b) => a - b);
    let ans: number = 1 << 30;
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        let j = i + 1;
        let k = n - 1;
        while (j < k) {
            const t: number = nums[i] + nums[j] + nums[k];
            if (t === target) {
                return t;
            }
            if (Math.abs(t - target) < Math.abs(ans - target)) {
                ans = t;
            }
            if (t > target) {
                --k;
            } else {
                ++j;
            }
        }
    }
    return ans;
}

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