leetcode 16. 3Sum Closest
2017-11-01 00:20
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这题和我上一道题类似,给定一个数组和一个目标值,在数组中取三个元素,使得这三个元素的和与目标值差距最小。借鉴上一道题3 Sum,我们先将数组进行了排序操作,让数组形成一个增序的数列,这样每取头尾两个元素,要使和增加则左边的游标加一,使和减小则右边的游标减一。这样做可以简化对满足条件的元素的查找。在这道题中,在判断数组元素个数不小于三之后,我们依旧可以先将数组排序,假定最终结果ans为前三个元素之和,然后下标i从0开始,一直到nums.size()-2,左边游标lo为i + 1,右边游标nums.size() - 1,重复下面的操作:
记nums[i]+nums[i+1]+nums[i+2]为sum
比较sum和ans离目标值的差距,选取差距小的为ans
当sum比目标值大,说明元素之和应该减小,则hi减一
当sum比目标值小,说明元素之和应该增大,则lo加一
当sum等于目标值,说明得到等于target的最终解,返回之
若i从0到nums.size()-2的循环结束,返回记录了和目标值差距最小的三元素之和ans
以下是代码实现:
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { if (nums.size() < 3) { int temp = 0; for (int i = 0; i < nums.size(); i++) temp += nums[i]; return temp; } else { sort(nums.begin(), nums.end()); int ans = nums[0] + nums[1] + nums[2]; for (int i = 0; i < nums.size() - 2; i++) { int lo = i + 1, hi = nums.size() - 1; while (lo < hi) { int sum = nums[i] + nums[lo] + nums[hi]; if (abs(target - sum) < abs(target - ans)) { ans = sum; } if (sum > target) hi--; if (sum == target) return target; if (sum < target) lo++; } } return ans; } } };
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