算法分析与设计——LeetCode:16. 3Sum Closest

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;    }};