3Sum Closest

标签：array   two points   遍历   

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).

与3SUM不同的是 不一定是有与target匹配的值 而是要寻找最接近的 显然要全部遍历 与3sum思想差不多 将火速组排序后 先枚举第一个数 然后在它后面的区间里 用头尾指针确定剩余两数 代码如下：

public class Solution {
    public int threeSumClosest(int[] num, int target) {
        if(num.length<=2)return 0;
        int res=0;
        Arrays.sort(num);
        boolean flag=true;
        for(int i=0;i<num.length-2;i++){
            int left=i+1;
            int right=num.length-1;
            while(left<right){
                int sum=num[left]+num[right]+num[i];
                if(flag==true){
                    res=sum;
                    flag=false;
                }
                else{
                   if(Math.abs(sum-target)<Math.abs(res-target)){
                       res=sum;
                   }
                }
                if(sum<target){
                    left++;
                }
                if(sum>target){
                    right--;
                }
                if(sum==target){
                    return target;
                }
            }
        }
        return res;
    }
}

3Sum Closest

标签：array   two points   遍历   

原文地址：http://blog.csdn.net/u012734829/article/details/44060059