3Sum

描述

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

  • Elements in a triplet (a,b,c) must be in non-descending order. (ie, a≤b≤ca \leq b \leq ca≤b≤c)
  • The solution set must not contain duplicate triplets.
    For example, given array S = {-1 0 1 2 -1 -4}.

A solution set is:

  1. (-1, 0, 1)
  2. (-1, -1, 2)

分析

先排序，然后左右夹逼，复杂度 O(n2)O(n^2)。

这个方法可以推广到k-sum，先排序，然后做k-2次循环，在最内层循环左右夹逼，时间复杂度是 O(max{nlogn,nk−1})O(\max{n \log n, n^{k-1}})。

代码

  1. // 3Sum
  2. // 先排序，然后左右夹逼，注意跳过重复的数
  3. // Time Complexity: O(n^2)，Space Complexity: O(1)
  4. class Solution {
  5.     public:
  6.     vector<vector<int>> threeSum(vector<int>& nums) {
  7.         vector<vector<int>> result;
  8.         if (nums.size() < 3) return result;
  9.         sort(nums.begin(), nums.end());
  10.         const int target = 0;
  12.         auto last = nums.end();
  13.         for (auto i = nums.begin(); i < last-2; ++i) {
  14.             if (i > nums.begin() && *i == *(i-1)) continue;
  15.             auto j = i+1;
  16.             auto k = last-1;
  17.             while (j < k) {
  18.                 if (*i + *j + *k < target) {
  19.                     ++j;
  20.                     while(*j == *(j - 1) && j < k) ++j;
  21.                 } else if (*i + *j + *k > target) {
  22.                     --k;
  23.                     while(*k == *(k + 1) && j < k) --k;
  24.                 } else {
  25.                     result.push_back({ *i, *j, *k });
  26.                     ++j;
  27.                     --k;
  28.                     while(*j == *(j - 1) && j < k) ++j;
  29.                     while(*k == *(k + 1) && j < k) --k;
  30.                 }
  31.             }
  32.         }
  33.         return result;
  34.     }
  35. };

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