#16 3Sum Closest

最新推荐文章于 2020-02-05 07:22:17 发布

原创最新推荐文章于 2020-02-05 07:22:17 发布 · 280 阅读

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本文介绍了一种算法，用于从整数数组中找出三个数，使其和最接近给定的目标值。通过排序和双指针技巧，有效地解决了这一问题。

题目如下：

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).

代码如下：

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums); 
        int len=nums.length;
        int closet=nums[0]+nums[1]+nums[2];
        for(int i=0;i<len-2;i++){
            int l=i+1;//left
            int r=len-1;//right
            /*if(l==r-1){
                closet=Math.abs(closet-target)<Math.abs(nums[i]+nums[l]+nums[r]-target)?closet:nums[i]+nums[l]+nums[r];
                continue;
            }*/
            while(l<r-1){
                if(closet==target)return closet;
                int a=nums[i]+nums[l]+nums[r]-target;
                if(nums[l]+nums[r]<target-nums[i])
                    l++;
                else r--;
                int b=nums[i]+nums[l]+nums[r]-target;
                if(a*b<0){
                    int min=Math.abs(a)<Math.abs(b)?a:b;
                    closet=Math.abs(closet-target)<Math.abs(min)?closet:min+target;
                    continue;
                }
            }
            closet=Math.abs(closet-target)<Math.abs(nums[i]+nums[l]+nums[r]-target)?closet:nums[i]+nums[l]+nums[r];
            
        }
        return closet;
    }
}

public class Solution {
	public int threeSumClosest(int[] num, int target) {
		int min = Integer.MAX_VALUE;
		int result = 0;
 
		Arrays.sort(num);
 
		for (int i = 0; i < num.length; i++) {
			int j = i + 1;
			int k = num.length - 1;
			while (j < k) {
				int sum = num[i] + num[j] + num[k];
				int diff = Math.abs(sum - target);
 
				if(diff == 0) return sum;
 
				if (diff < min) {
					min = diff;
					result = sum;
				}
				if (sum <= target) {
					j++;
				} else {
					k--;
				}
			}
		}
 
		return result;
	}
}