Leetcode 16. 3Sum Closest

16. 3Sum Closest

• Total Accepted: 85780
• Total Submissions: 289017
• Difficulty: Medium

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).

思路：将序列升序排序后，设定ans等于a[0],a[1],a[2]，然后用三个指针集i，j，k遍历有序序列，sum=a[i]+a[j]+a[k]，如果sum>ans，就更新ans。每一轮特定i下关j，k都遍历完后，就比较ans是否等于target，如果等于就返回当前的a[i],a[j],a[k]。

原来还想排序后找到最小的>=target的数A，然后得到由这个数递减的最大的<=target的数B。比较AB和target的接近程度，然后返回相应结果。

代码：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int i,n=nums.size(),ans=0;
        if(n<=3){
            for(i=0;i<n;i++){
                ans+=nums[i];
            }
            return ans;
        }
        sort(nums.begin(),nums.end());
        ans=nums[0]+nums[1]+nums[2];
        int sum,j,k;
        for(i=0;i<n;i++){
            j=i+1;
            k=n-1;
            while(j<k){
                sum=nums[i]+nums[j]+nums[k];
                if(abs(target-ans)>abs(target-sum)){
                    ans=sum;
                }
                sum>target?k--:j++;
            }
            if(ans==target) return ans;//优化
        }
        return ans;
    }
};