【leetcode】Array——3Sum Closest（16）_iven an array a1,a2...an, you have to print the-优快云博客

本文链接：https://blog.youkuaiyun.com/u013127687/article/details/51004054

本文介绍了一种优化算法，用于解决在给定整数数组中找到三个数使其和最接近指定目标值的问题。通过双指针法改进原始的暴力搜索，将时间复杂度从O(n^3)优化到O(n^2)，并提供了具体实现代码。

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题目：

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).

思路：暴力遍历，找差值最小的返回。时间复杂度是O(n^3)

代码：略

优化：遍历第一个数，后面两个数用two pointers从两端往中间遍历，时间复杂度O(n^2)

代码：

public int threeSumClosest(int[] nums, int target) {
	Arrays.sort(nums);
	int minDif=Integer.MAX_VALUE;
	int res=0;
	for(int i=0;i<nums.length;i++){
		int start = i+1;
		int end = nums.length-1;
		while(start<end){
			int sum = nums[i]+nums[start]+nums[end];
			if(sum==target)
				return target;
			else if(sum<target){
				start++;
			}else{
				end--;
			}
			if(Math.abs(sum-target)<Math.abs(minDif)){
				minDif=sum-target;
				res=sum;
			}
				
		}
	}
	return res;
}

或者如下（下面方法在leetcode runtime比上面短很多，虽然算法原理一样，并不知道为啥~~~）

public int threeSumClosest(int[] nums, int target) {
	Arrays.sort(nums);
	int minDif=Integer.MAX_VALUE;
	for(int i=0;i<nums.length-2;i++){
		int curDif=Two(nums,nums[i],i+1,target-nums[i]);
		if(curDif==0)
			return target;
		if(Math.abs(curDif)<Math.abs(minDif)){
			minDif=curDif;
		}
	}
	
	return target+minDif;
}
private int Two(int []nums,int z,int start,int target){
	int minDif=Integer.MAX_VALUE;
	int end = nums.length-1;
	while(start<end){
		int sum = nums[start]+nums[end];
		if(sum==target)
			return 0;
		else if(sum>target){
			end--;
		}else{
			start++;
		}
		if(Math.abs(sum-target)<Math.abs(minDif))
			minDif=sum-target;
	}
	return minDif;
}