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. public class Solution {
  5.     public List<List<Integer>> threeSum(int[] nums) {
  6.         List<List<Integer>> result = new ArrayList<>();
  7.         if (nums.length < 3) return result;
  8.         Arrays.sort(nums);
  9.         final int target = 0;
  11.         for (int i = 0; i < nums.length - 2; ++i) {
  12.             if (i > 0 && nums[i] == nums[i-1]) continue;
  13.             int j = i+1;
  14.             int k = nums.length-1;
  15.             while (j < k) {
  16.                 if (nums[i] + nums[j] + nums[k] < target) {
  17.                     ++j;
  18.                     while(nums[j] == nums[j-1] && j < k) ++j;
  19.                 } else if(nums[i] + nums[j] + nums[k] > target) {
  20.                     --k;
  21.                     while(nums[k] == nums[k+1] && j < k) --k;
  22.                 } else {
  23.                     result.add(Arrays.asList(nums[i], nums[j], nums[k]));
  24.                     ++j;
  25.                     --k;
  26.                     while(nums[j] == nums[j-1] && j < k) ++j;
  27.                     while(nums[k] == nums[k+1] && j < k) --k;
  28.                 }
  29.             }
  30.         }
  31.         return result;
  32.     }
  33. };

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