3Sum Closest(最接近target的三个数字的和)leetcode16

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).

code1：（暴力破解）

 public static int threeSumClosest(int[] nums, int target) {
    	int closest=Integer.MAX_VALUE;
    	int resultSum=0;
        Arrays.sort(nums);
        for(int i=0;i<nums.length;i++)
         for(int j=i+1;j<nums.length;j++)
             for(int k=j+1;k<nums.length;k++)
                 if(Math.abs(target-nums[i]-nums[j]-nums[k])<closest) {
                     closest=Math.abs(target-nums[i]-nums[j]-nums[k]);
                     resultSum=nums[i]+nums[j]+nums[k];
                     //System.out.println(closest);
                    }
        return resultSum;
}

code2：（优化后的）

 public static int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int result=nums[0]+nums[1]+nums[2];
        for(int i=0;i<nums.length-2;){
            int front=i+1;
            int last=nums.length-1;
            while(front<last){
                int sum=nums[i]+nums[front]+nums[last];
                if(sum==target) return sum;
                if(Math.abs(sum-target)<Math.abs(target-result))
                    result=sum;
                if(sum>target){
                    last--;
                     while(last>front&&nums[last]==nums[last+1]) last--;
                }
                if(sum<target){
                    front++;
                    while(front<last&&nums[front]==nums[front-1]) front++;
                }
                
               
            }
            i++;
            while(i<nums.length&&nums[i]==nums[i-1])
                i++;
        }
        return result;
    }
}