[Leetcode] 3Sum Closest

使用2sum降低复杂度为O（N^2)

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int closest =Integer.MAX_VALUE;
        boolean first = true;//if the first time, then there is no need to compare the tmpsum with the closest
        for(int i=0;i<nums.length-2;i++){
            int num = nums[i];
            int left = i+1;
            int right = nums.length-1;
            
            while(left<right){
                int tmpsum = num + nums[left] + nums[right];
                if(first){
                    closest = tmpsum;
                    first =false;
                }else{
                    if(Math.abs(tmpsum-target)<Math.abs(closest - target)){
                        closest = tmpsum;
                    }
                }
                //move the left or the right pointer according to whether the tmpsum is larger than the target
                if(tmpsum<target){
                    left++;
                }
                else if(tmpsum>target){
                    right--;
                }else{
                    right--;
                    left++;
                }
            }
        }
        return closest;
    }
}

 

这个问题几乎和 3Sum 是一样的，我们但是这里是要找最为接近的三元组，所以选择的标准不一样而已，其他的一样。

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int mindiff=Integer.MAX_VALUE;
        int closestsum=0;
        for(int i=0;i<nums.length;i++){
            int num1=nums[i];
            if(i>0&&num1==nums[i-1]){
                continue;
            }
            for(int j=i+1;j<nums.length;j++){
                int num2=nums[j];
                if(j>i+1&&num2==nums[j-1]){
                    continue;
                }
                for(int k=j+1;k<nums.length;k++){
                    int num3=nums[k];
                    if(k>j+1&&num3==nums[k-1]){
                        continue;
                    }
                    int sum=num1+num2+num3;
                    int diff=Math.abs(sum-target);
                    if(diff<mindiff){
                        mindiff=diff;
                        closestsum=sum;
                        if(diff==0) return sum;
                    }
                    if(sum>target){
                        break;
                    }
                }
            }
        }
        return closestsum;
    }
}