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;
    }
    
};