【leetcode-16】3Sum Closest(java)

最新推荐文章于 2024-04-29 23:48:08 发布

zdavb

最新推荐文章于 2024-04-29 23:48:08 发布

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本文解决了一个与LeetCode-15相似的问题，但目标是找到数组中三个整数的和，使得这个和与给定的目标值之间的差最小。通过排序数组并使用双指针法，实现了一个时间复杂度为O(n^2)的解决方案。

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问题描述：
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).

分析：该题与leetcode-15及其类似，只是将和为0条件改为了与目标值最小。也即减去目标值的差最接近0.

代码如下：348ms，时间复杂度：O(n2)

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int left,right;
        Arrays.sort(nums);//排序
        int sum = Integer.MAX_VALUE;
        int minusWithTarget = Integer.MAX_VALUE;
        int sumWithTriple;
        int minus;

        for(int i = 0;i<nums.length;i++){
            if(i-1>=0&&nums[i]==nums[i-1])
                continue;
            left = i+1;right = nums.length-1;
            while(left<right){
                sumWithTriple = nums[i]+nums[left]+nums[right];
                minus = sumWithTriple-target;
                if(minus==0){
                    return sumWithTriple;
                }else if(minus>0){
                    if(minus<minusWithTarget){
                        minusWithTarget = minus;
                        sum = sumWithTriple;
                    }
                    right--;
                }else {
                    if(-minus<minusWithTarget){
                        minusWithTarget = -minus;
                        sum=sumWithTriple;
                    }
                    left++;
                }
            }
        }
        return sum;
    }
}