[LeetCode] 16. 3Sum Closest

最新推荐文章于 2021-08-11 08:50:42 发布

Ber03

最新推荐文章于 2021-08-11 08:50:42 发布

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分类专栏： LeetCode 文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/Bertram03/article/details/86763181

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本文深入解析LeetCode上3SumClosest问题的算法实现，通过先固定一个数，使用二分查找法寻找另外两个数，以求得最接近目标值的三数之和。文章详细介绍了算法步骤，并附带了完整的Java代码实现。

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原题链接： https://leetcode.com/problems/3sum-closest/

1. 题目描述

Given an array nums of n integers 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:

Given array nums = [-1, 2, 1, -4], and target = 1.

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

2. 做题思路

同3Sum，先找到a，后用二分去找b、c
记录最小的差，如果遇到更小的差，把差替换一下。

代码

class Solution {
   public int threeSumClosest(int[] nums, int target) {
			int length = nums.length;
			int cha = 9999999;
			int res = 0;
			Arrays.sort(nums);
			for(int i = 0;i<length-2;i++) {
				if(i>0 && nums[i] == nums[i-1]) {
					continue;
				}
				int a = erfen(i,nums,target);
				int temp = Math.abs(a-target);
				if(temp <= cha) {
					cha = temp;
					res = a;
				}
					
			}
			 
			 return res;
		 }
		 
		public int erfen(int index, int[] nums,int tar) {
			int l = index +1;
			int r = nums.length-1;
			
			int cha=9999999;
			int closedsum = 0;
			int sum = 0;
			
			while(l<r) {
				sum = nums[l]+nums[r]+nums[index];
				if( sum == tar) {
					closedsum = tar;
	                l++;
	                r--;
					return closedsum;
					
				}
				else if(sum <tar) {
					if(tar- sum < cha) {
					cha = tar-sum;
					closedsum = sum;
					}
					l++;
					
				}
				else if(sum > tar) {
					if(sum-tar < cha) {
					cha =  sum-tar;
					closedsum = sum;
						}
					r--;
				}
			}
			
			return closedsum;
		}
	 
}