leetcode 3Sum Closest题解

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).

中文理解：

给定一个数和数组，从数组中找到三个元素的和距离给定数最近的数，返回。

解题思路：

首先对数组进行排序，然后外层循环从0到nums.length-2，内层循环则设置两个指针start,end，如果nums[i]+nums[start]+nums[end]大于target则end--，同时比较距离target的距离是否小了，如果nums[i]+nums[start]+nums[end]小于target，则start++,同时比较距离target的距离是否小了，小了更新，如果nums[i]+nums[start]+nums[end]等于target，返回target。

代码（java）：

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if(nums.length<=3){
            int sum=0;
            for(int val:nums){
                sum+=val;
            }
            return sum;
        }
        Arrays.sort(nums);
        int res=nums[0]+nums[1]+nums[2];
        for(int i=0;i<nums.length-2;i++){
            int start=i+1,end=nums.length-1;
            while(start<end){
                if(target<nums[i]+nums[start]+nums[end]){
                    if(Math.abs(nums[i]+nums[start]+nums[end]-target)<Math.abs(res-target))res=nums[i]+nums[start]+nums[end];
                    end--;
                }
                else if(target>nums[i]+nums[start]+nums[end]){
                    if(Math.abs(nums[i]+nums[start]+nums[end]-target)<Math.abs(res-target))res=nums[i]+nums[start]+nums[end];
                    start++;
                }
                else if(target==nums[i]+nums[start]+nums[end])return target;
            }
        }
        return res;
    }
}