LeetCode – Refresh – 3sumclosest

csid_502

于 2015-03-18 06:59:00 发布

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原文链接：http://www.cnblogs.com/shuashuashua/p/4346150.html

本文介绍了一个类似于三数之和的问题解决方法，并提供了一段C++代码实现。该算法通过先排序再使用双指针技巧来寻找数组中三个元素的组合，使得这三个数的和最接近给定的目标值。

This problem is almost like 3sum. Actually it needs the same skip steps for j and k. But LeetCode did not put the test case there. So just put like this.

 1 class Solution {
 2 public:
 3     int threeSumClosest(vector<int> &num, int target) {
 4         int len = num.size(), result = num[0] + num[1] + num[2], i = 0, j = 0, k = 0;
 5         sort(num.begin(), num.end());
 6         for (i = 0; i < len-2; i++) {
 7             if (i > 0 && num[i] == num[i-1]) continue;
 8             j = i+1, k = len-1;
 9             while (j < k) {
10                 int sum = num[i] + num[j] + num[k];
11                 if (fabs(sum - target) < fabs(result - target)) result = sum;
12                 if (sum > target) k--;
13                 else j++;
14             }
15         }
16         return result;
17     }
18 };