LeetCode[16]: 3Sum Closest
2018-03-10 10:59
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题目描述
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
这道题和之前的两数只和的解决方式类似,区别是在其基础上增加了第三个数。我们可以通过固定其中一个数numsinumsi,然后寻找两数之和最接近target−numsitarget−numsi,将问题转化为两数之和的问题。首先将数据排序,遍历数组固定numsinumsi然后使j=i+1,k=length−1j=i+1,k=length−1,通过移动j,kj,k两个下标,来使numsj+numsk逼近target−numsinumsj+numsk逼近target−numsi。然后保存当前最优解和当前三数之和更接近目标target的值。代码
class Solution { public int threeSumClosest(int[] nums, int target) { int threeSum = nums[0] + nums[1] + nums[2]; int closestNum = threeSum; //先排序 Arrays.sort(nums); for (int i=0; i < nums.length-2; i++) { int twoSum = target - nums[i]; int j = i + 1, k = nums.length-1; while (j < k) { threeSum = nums[i] + nums[j] + nums[k]; if(Math.abs(threeSum-target) < Math.abs(closestNum-target)) closestNum = threeSum; if (nums[j] + nums[k] < twoSum) { j++; } else if (nums[j] + nums[k] > twoSum) { k--; } else { return target; } } } return closestNum; } }
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