3Sum Closest

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).

分析

先排序，然后左右夹逼，复杂度 O(n2)O(n^2)。

代码

  1. // 3Sum Closest
  2. // 先排序，然后左右夹逼//
  3. // Time Complexity: O(n^2), Space Complexity: O(1)
  4. class Solution {
  5. public:
  6.     int threeSumClosest(vector<int>& nums, int target) {
  7.         int result = 0;
  8.         int min_gap = INT_MAX;
  10.         sort(nums.begin(), nums.end());
  12.         for (auto a = nums.begin(); a != prev(nums.end(), 2); ++a) {
  13.             auto b = next(a);
  14.             auto c = prev(nums.end());
  16.             while (b < c) {
  17.                 const int sum = *a + *b + *c;
  18.                 const int gap = abs(sum - target);
  20.                 if (gap < min_gap) {
  21.                     result = sum;
  22.                     min_gap = gap;
  23.                 }
  25.                 if (sum < target) ++b;
  26.                 else              --c;
  27.             }
  28.         }
  30.         return result;
  31.     }
  32. };

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/linear-list/array/3sum-closest.html