【leetcode】Array——3Sum Closest（16）

題目：

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：暴力遍歷，找差值最小的返回。時間複雜度是O(n^3)

代碼：略

優化：遍歷第一個數，後面兩個數用two pointers從兩端往中間遍歷，時間複雜度O(n^2)

代碼：

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，雖然算法原理一樣，並不知道爲啥~~~）

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;
}