16.3Sum Closest (Two-Pointers)
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>& nums, int target) { int size = nums.size(); sort(nums.begin(), nums.end()); diff = INT_MAX; for(int i = 0; i < size-2; i++){ if(i>0 && nums[i]==nums[i-1]) continue; find(nums, i+1, size-1, target-nums[i]); if(diff == 0) return target; } return target+diff; } void find(vector<int>& nums, int start, int end, int target){ int sum; while(start<end){ sum = nums[start]+nums[end]; if(sum == target){ diff = 0; return; } else if(sum>target){ do{ end--; }while(end!=start && nums[end] == nums[end+1]); if(sum-target < abs(diff)) diff = sum - target; } else{ do{ start++; }while(start!= end && nums[start] == nums[start-1]); if(target - sum < abs(diff)) diff = sum - target; //不能只在最后检查:可能会有这种情况,前一次sum>target,这次sum<target,而且下次就start==end,那么很可能前一次的sum比这次的sum更接近target } } } private: int diff; //how much bigger is the sum };