Leetcode-16. 3Sum Closest

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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).

这个算是3Sum的变体，可以用两端逼近来做，复杂度还是O（n^2），不过不知道为何我的效率比较低。Your runtime beats 4.08% of java submissions.

public class Solution {
public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
if(nums.length == 3) return nums[0] + nums[1] + nums[2];
int miniDistance = Integer.MAX_VALUE;

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

if (i == 0 || (i > 0 && nums[i] != nums[i-1])) {
int l = i + 1,r = nums.length - 1;
int lastDistance = nums[l]+nums[r] + nums[i]-target;
while(l<r){
if(Math.abs(nums[l]+nums[r - 1] + nums[i] - target) <= Math.abs(lastDistance) && r - l > 1){
r --;
lastDistance = nums[l]+nums[r] + nums[i]-target;
}
else if( Math.abs(nums[l+1]+nums[r] + nums[i] - target) <= Math.abs(lastDistance) && r - l > 1){
l ++;
lastDistance = nums[l]+nums[r] + nums[i]-target;
}else{
r--;
l++;
}
}
miniDistance = Math.abs(miniDistance) > Math.abs(lastDistance) ? lastDistance : miniDistance;
}
}
return miniDistance + target;
}
}