[Leetcode] 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).

跟3SUM一样的想法，排序数组，先选一个数出来，再用target减去这个数，然后用找和为给定值的算法找最小差值，时间复杂度为O(n^2) , 要比枚举所有情况的O(n^3)来的快！

 1 class Solution {

 2 public:

 3     int getDiff(vector<int> &num, int idx, int target) {

 4         int low = idx, high = num.size() - 1;

 5         int res = INT_MAX, tmp;

 6         while (low < high) {

 7             tmp = num[low] + num[high] - target;

 8             res = abs(res) > abs(tmp) ? tmp : res;

 9             if (tmp > 0) --high;

10             else if (tmp < 0) ++low;

11             else break;

12         }

13         return res;

14     }

15     

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

17         int newtarget;

18         int diff = INT_MAX, tmp;

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

20         for (int i = 0; i < num.size() - 2; ++i) {

21             newtarget = target - num[i];

22             tmp = getDiff(num, i + 1, newtarget);

23             diff = abs(diff) > abs(tmp) ? tmp : diff;

24         }

25         return target + diff;

26     }

27 };