16. 3Sum Closest
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).
在给定数列中找出三个数,使和最接近 target
还是先排序,再左右夹逼
public class Solution { public int threeSumClosest(int[] num, int target) { Arrays.sort(num); int ret = num[0] + num[1] + num[2]; int len = num.length; for (int i = 0; i <= len - 3; i++) { // first number : num[i] int j = i + 1; // second number int k = len - 1; // third number while (j < k) { int sum = num[i] + num[j] + num[k]; if (Math.abs(sum - target) < Math.abs(ret - target)) ret = sum; if (sum < target) { ++j; } else if (sum > target) { --k; } else { ++j; --k; } } } return ret; } }
题意:数组中每三个元素进行求和,找出所有和中大小最接近target的和,并返回这个和与target之间的差值。
解题思路:使用一个变量mindiff来监测和与target之间的差值,如果差值为0,直接返回sum值。
代码:
class Solution: # @return an integer def threeSumClosest(self, num, target): num.sort() mindiff=100000 res=0 for i in range(len(num)): left=i+1; right=len(num)-1 while left<right: sum=num[i]+num[left]+num[right] diff=abs(sum-target) if diff<mindiff: mindiff=diff; res=sum if sum==target: return sum elif sum<target: left+=1 else: right-=1 return res