16 3sum closest

16 3sum closest

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).

想法1：

    和3sum的想法类似，先对数组进行排序，然后给closestNum赋初值等于nums[0]+nums[1]+nums[2]。head，tail分别指向数组的头和尾，如果sum>target,则tail--;否则，head++；如果(sum-target)的绝对值小于(closestNum-target)的绝对值，那么更新closestNum=sum。

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int head, tail, closestNum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size(); i++){
            head = i + 1;
            tail = nums.size() - 1;
            int sum;
            bool isCalculate = false;
            while (head < tail){
                sum = nums[head] + nums[tail] + nums[i];
                if (sum<target){
                    if (abs(target - sum)<abs(target - closestNum))
                        closestNum = sum;
                    head++;
                }
                else if (nums[head] + nums[tail] + nums[i]>target){
                    if (abs(target - sum) < abs(target - closestNum))
                        closestNum = sum;
                    tail--;
                }
                else {
                    return target;
                }
            }
        }
        return closestNum;
    }
};