leetcode 16. 3Sum Closest

最新推荐文章于 2020-03-26 11:43:19 发布

原创最新推荐文章于 2020-03-26 11:43:19 发布 · 308 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

LeetCode 专栏收录该内容

28 篇文章

订阅专栏

本文介绍了一种寻找数组中三个整数之和最接近给定目标值的有效算法。通过先排序再使用双指针技术，实现了从立方级别的时间复杂度降至平方级别。文章详细解释了算法步骤，并附带完整的C++实现代码。

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).

解：这一题需要先进行排序，这样就可以将三方的复杂度降低到平方的复杂度。假定一个id1后遍历后面的数组，然后后移id1.

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        if(nums.size() < 3) return 0;
        int closest = nums[0]+nums[1]+nums[2];
        std::sort(nums.begin(), nums.end());
        for(int first = 0 ; first < nums.size()-2 ; ++first) {
            if(first > 0 && nums[first] == nums[first-1]) continue;
            int second = first+1;
            int third = nums.size()-1;            
            while(second < third) {
                int curSum = nums[first]+nums[second]+nums[third];
                if(curSum == target) return curSum;
                if(abs(target-curSum)<abs(target-closest)) {
                    closest = curSum;
                }
                if(curSum > target) {
                    --third;
                } else {
                    ++second;
                }
            }
        }
        return closest;
    }
};