LeetCode 3Sum Closest

class Solution {

public:

    int threeSumClosest(vector<int> &num, int target) {

        int len = num.size();

        if (len < 3) {

            return 0;

        }

        sort(num.begin(), num.end());

        

        int sum = num[0] + num[1] + num[2];

        for (int i=0; i<len-2; i++) {

            int p = i + 1;

            int q = len - 1;

            while (p < q) {

                int csum = num[i] + num[p] + num[q];

                if (abs(target - csum) < abs(target - sum)) {

                    sum = csum;

                }

                if (csum > target) {

                    q--;

                } else if (csum < target) {

                    p++;

                } else {

                    return csum;

                }

            }

        }

        return sum;

    }

};

n方时间复杂度

 

第二轮：

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> &num, int target) {

        sort(num.begin(), num.end());

        int len = num.size();

        if (len < 3) {

            return 0;

        }

        int min_diff = INT_MAX;

        int closest = 0;

        for (int i=0; i<len; i++) {

            int rest = target - num[i];

            int p = i + 1;

            int q = len - 1;

            while (p < q) {

                int t = num[p] + num[q];

                int tdiff = abs(t - rest);

                if (tdiff < min_diff) {

                    closest = t + num[i];

                    min_diff = tdiff;

                }

                if (t > rest) {

                    q--;

                } else if (t < rest) {

                    p++;

                } else {

                    break;

                }

            }

        }

        return closest;

    }

    

};