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).
题解:3Sum类似,只要维护一个变量closest作为当前和target最近的和。如果之间能够凑出来target,返回target就可以了。
有一个坑就是closest初始化的时候不能初始化为Integer.MAXVALUE,否则closest-target当target<0的时候会溢出。
比如输入 [-3,-2,-5,3,-4], -1 Integer.MAXVALUE-(-1)瞬间就溢出了。
代码如下:
1 public class Solution { 2 public int threeSumClosest(int[] num, int target) { 3 if(num == null || num.length == 0) 4 return 0; 5 Arrays.sort(num); 6 int closest = Integer.MAX_VALUE/2; 7 8 for(int i = 0;i < num.length;i++){ 9 int start = i + 1; 10 int last = num.length-1; 11 while(start < last){ 12 int sum = num[i] + num[start] + num[last]; 13 if(sum == target) 14 return sum; 15 else if(sum < target) 16 { 17 start++; 18 } 19 else{ 20 last--; 21 } 22 closest = Math.abs(closest-target) > Math.abs(sum - target)?sum:closest; 23 } 24 } 25 26 return closest; 27 } 28 }