算法内容
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) 。
检查例子：
输入：1,1,1,0 100
输入：1,1,1,0 -100
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/3sum-closest
算法思想
最接近三数之和思想与三数之和思想相似，只不过在一些操作细节上需要注意一些。如下是该算法相较于三数之和算法的区别点：
//count_num1记录三数之和的count_num与目标值之间的距离。
count_num1=count_num-target;
//如果|count_num1|小于|count_3-target|则count_num=count_num否则count_num=count_3。
if (count_num1<0){
count_num =-count_num1 < -(count_3-target) ? count_num:count_3;
}else{
count_num =count_num1 < (-(count_3-target)) ? count_num:count_3;
}
整体算法
- 双指针
public class Day_21 {
static Scanner input=new Scanner(System.in);
public static int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int count=nums.length-2;
int count_num=10000;
int count_num1=0;
for (int i = 0; i < count; i++) {
if (i>0&&nums[i]==nums[i-1]){
continue;
}
int y=i+1;
for (int j = nums.length-1; j >y ; ) {
if (j<nums.length-1&&nums[j]==nums[j+1]){
j--;
continue;
}
int count_3=nums[i]+nums[j]+nums[y];
if (count_3<target){
y++;
count_num1=count_num-target;
if (count_num1<0){
count_num =-count_num1 < -(count_3-target) ? count_num:count_3;
}else{
count_num =count_num1 < (-(count_3-target)) ? count_num:count_3;
}
}
if (count_3>target){
j--;
count_num1=count_num-target;
if (count_num1<0){
count_num =-count_num1 < count_3-target ? count_num:count_3;
}else{
count_num =count_num1 < count_3-target ? count_num:count_3;
}
}
if (count_3==target){
return count_3;
}
}
}
return count_num;
}
public static void main(String[] args) {
System.out.print("请输入数组大小：");
int num=input.nextInt();
int [] nums=new int[num];
System.out.println("请输入数组数据：");
for (int i = 0; i < num; i++) {
System.out.println("数字"+i+":");
nums[i]=input.nextInt();
}
System.out.print("输入目标值：");
int target=input.nextInt();
System.out.println(threeSumClosest(nums,target));
}
}
尾语
以上属于个人见解，有好的想法可以在下方评论写出自己的想法，大家一起进步。该题是力扣上的题，若有侵权，请及时告知。该题链接：https://leetcode-cn.com/problems/3sum-closest