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This problem is very similar to 3Sum. You only need to maintain a variable for the sum that is closet to target. Also, some corner cases need to be handled; for example, nums does not have more than 2 elements.
The code is as follows, which is quite self-explanatory.
1 int threeSumClosest(vector<int>& nums, int target) { 2 sort(nums.begin(), nums.end()); 3 while (nums.size() <= 2) 4 return accumulate(nums.begin(), nums.end(), 0); 5 int ans = nums[0] + nums[1] + nums[2]; 6 for (int i = 0; i < nums.size() - 2; i++) { 7 int left = i + 1, right = nums.size() - 1; 8 while (left < right) { 9 int temp = nums[i] + nums[left] + nums[right]; 10 if (abs(temp - target) < abs(ans - target)) 11 ans = temp; 12 if (temp == target) return ans; 13 if (temp > target) right--; 14 else left++; 15 } 16 } 17 return ans; 18 }
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4567740.html
