标签：find input evel else get mos ble begin style
Problem statement:
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).
Solution:
It is a change for 15. 3Sum, but quite similar, this problem requires to return the sum of three numbers which is most close to the target.
Just do the absolute value test when we update the index, left and right, keep the closest sum and return at the end or in the loop.
Time complexity, two level loops, O(n * n).
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); int closest_sum = nums[0] + nums[1] + nums[2]; int size = nums.size(); for(int i = 0; i < size - 1; i++){ int left = i + 1; int right = size - 1; while(left < right){ if(abs(nums[i] + nums[left] + nums[right] - target) < abs(closest_sum - target)){ closest_sum = nums[i] + nums[left] + nums[right]; } if(nums[i] + nums[left] + nums[right] == target){ return target; } else if (nums[i] + nums[left] + nums[right] > target){ right--; } else { left++; } } } return closest_sum; } };
标签：find input evel else get mos ble begin style
原文地址：http://www.cnblogs.com/wdw828/p/6883080.html
