3Sum closest

【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).

【解析】

和 3Sum解题报告 很像，与之不同的是，不再是求三个数的和是不是为0，而是看三个数的和与target的差是否为最小，只需记录当前最优解并不断更新其值就可。

【Java代码】O(n^2)

[java]  view plain  copy

1. public class Solution {  
2.     public int threeSumClosest(int[] num, int target) {  
3.         if (num == null || num.length < 3) return 0;  
4.           
5.         Arrays.sort(num);  
6.           
7.         int ret = 0;  
8.         int closestDist = Integer.MAX_VALUE;  
9.         int len =  num.length;  
10.         for (int i = 0; i < len-2; i++) {  
11.             if (i > 0 && num[i] == num[i-1]) continue;  
12.               
13.             int l = i+1, r = len-1;  
14.             while (l < r) {  
15.                 int sum = num[i] + num[l] + num[r];  
16.                 if (sum < target) {  
17.                     if (target-sum < closestDist) {  
18.                         closestDist = target - sum;  
19.                         ret = sum;  
20.                     }  
21.                     l++;  
22.                 } else if (sum > target) {  
23.                     if (sum-target < closestDist) {  
24.                         closestDist = sum - target;  
25.                         ret = sum;  
26.                     }  
27.                     r--;  
28.                 } else { //when sum == target, return sum.  
29.                     return sum;  
30.                 }  
31.             }  
32.         }  
33.           
34.         return ret;  
35.     }  
36. }  

容易出错的地方是，把 ret 初始值设为 Integer.MAX_VALUE，然后后面计算 closestDist = Math.abs(ret - target)，这样会导致溢出！！