算法分析与设计——LeetCode:16. 3Sum Closest_算法分析16.3.2-优快云博客

本文链接：https://blog.youkuaiyun.com/sysu_CN/article/details/78428825

本文介绍了一种高效算法，用于寻找数组中三个整数的组合，使其和最接近给定的目标值。通过先排序再使用双指针技术，实现了O(n^2)的时间复杂度。

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题目

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).

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        return closest;
    }
};

思路

如果遍历所有情况，复杂度将为O(n^3)。为了降低其复杂度，先按大小将序列进行排序，调用sort函数，复杂度为O(nlog2n)，然后如下图

横条代表排序后的序列，first从头到尾遍历一遍，每一轮中，second指向first+1，third指向序列最后一个数，若和大于target，则将third前移一位，使其变小，若和小于target，则将second后移一位，这样最后就可以找到与target最接近的和，算法复杂度为O(n^2)。

代码

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