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[Leetcode] 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.

  Array

体会:

1. 这道题最直接的想法肯定是想 A3n的那种解法,就是遍历所有可能的三个数的排列,然后找结果。可是这样就是A3n = n(n-1)(n-2) = O(n3)的解法,会exceed time limit。

2. 然后就是低于3次的解法了。

 

class Solution {    public:        int threeSumClosest(vector<int> &num, int target) {            // we don‘t want to change the original vector probably            // how to declare vector in C++            vector<int> v(num.begin(), num.end());                        int result = 0;            int size = v.size();            if (size <= 3) {                for (int i = 0; i < size; i++){                    result += v[i];                    }                return result;            }            std::sort(v.begin(), v.end());            result = v[0] + v[1] + v[2];            int sum = 0;            for (int i = 0; i < size - 2; i++) {                int j = i + 1;                int k = size - 1;                while (j < k) {                    sum = v[i] + v[j] + v[k];                    if (std::abs(target - sum) < std::abs(target - result)) {                        result = sum;                        if (result == target) {                            return result;                        }                    }                    if (sum > target) {                        k--;                    } else {                        j++;                    }                }            }            return result;        }};

 

[Leetcode] 3Sum Closest