题解:3Sum Closest
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题目:
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).
解题思路:首先将该向量排序。之后,设置first, second, third 三个坐标,first自然从最小也就是0开始,second随其后,third从最后开始往前遍历。用一个大循环套一个小循环,大循环是first的递增,小循环中,如果检测到三个数的和比target大,那么third的数值减小,否则second增加,知道判断second > third时跳出小循环。就这样往复得到result.
代码如下:
int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); int result = nums[0] + nums[1] + nums[2]; for(int first = 0; first < nums.size() - 2; first++) { int second = first + 1, third = nums.size() - 1; while(second < third) { int num = nums[first] + nums[second] + nums[third]; if(abs(num - target) < abs(result - target))result = num; if(num < target)second++; elsethird--; } } return result;}
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