Three Sum Closest || LeetCode-16

Problem link： https://leetcode.com/problems/3sum-closest/#/description

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.

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

<问题描述>

给定一个整形数组，需要从数组中找出三个整数，使得它们的和最接近另外一个给定的整数target

你不能使用同一个元素两次

给出的threeSumClosest函数应该返回最接近target的整数值

输入：numbers = [-1 2 1 -4], target = 1 (注意：输入的数组不一定有序)

输出：2 (-1 + 2 + 1 = 2)

Approach #Solution (Sorting With Two Pointers)

Analysis

方法和Three Sum问题的差不多，只不过现在是要输出一个整数即可，并且不需要考虑nums[i]、nums[low]、nums[high]元素的重复问题，显得更加简单。需要注意一点的就是，判断low和high指针如何移动的条件。

Code

public int threeSumClosest(int[] nums, int target){
    int closest_target = nums[0] + nums[1] + nums[2]; // closest_target为最接近的3Sum
    int smallest_difference_value = Math.abs(closest_target - target); // 3Sum与target之间最小的差值
    Arrays.sort(nums);

    for (int i = 0; i < nums.length - 2; i++) {
        int low = i + 1;
        int high = nums.length - 1;

        while(low < high){
            int sum = nums[i] + nums[low] + nums[high];
            int temp = Math.abs(sum - target);
            if (temp < smallest_difference_value) {
                closest_target = sum;
                smallest_difference_value = temp;
            }
            // sum的值必须无限接近target，以此来判断low和high指针的移动
            if (sum < target) {
                low++;
            }else
                high--;
        }
    }
    return closest_target;
}

Complexity

时间复杂度：O(n^2)

空间复杂度：O(1)取决于排序算法