leetcodeに3Sumアルゴリズムの問​​題があり、時間の複雑さを軽減する方法



There Is 3sum Algorithm Problem Leetcode

今日、私は誤ってレフトコードで3Sumトピックを作成し、それが非常に難しいことに気づきました。

トピック:配列が与えられた場合、合計が0の配列を3つ見つけて、繰り返されないすべての結果のコレクションを返します。



Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: The solution set must not contain duplicate triplets. Example: Given array nums = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]

私の頭の中の最初の解決策:

// Convenience the entire array, find all the combinations, then compare them one by one, time complexity O(n^3) public List<List<Integer>> threeSum(int[] nums) { List<List<Integer>> solutionList = new ArrayList<List<Integer>>() if (nums != null && nums.length >2) { for (int i=0i<nums.length-2i++) { for (int j=i+1j<nums.length-1j++) { for (int k=j+1k<nums.lengthk++) { int sum = nums[i] + nums[j] + nums[k] if (sum == 0) { List<Integer> list = new ArrayList<Integer>() list.add(nums[i]) list.add(nums[j]) list.add(nums[k]) // sort the list List<Integer> sorted = bubbleSort(list) //System.out.println(sorted) if (solutionList.size() >0) { if (!solutionList.contains(sorted)) { solutionList.add(sorted) } } else { solutionList.add(sorted) } } } } } } return solutionList } public List<Integer> bubbleSort(List<Integer> list) { for (int i=0i<list.size() -1i++) { for (int j=i+1j<list.size()j++) { if (list.get(i) > list.get(j)) { Integer temp = list.get(i) list.set(i,list.get(j)) list.set(j, temp) } } } return list }

このように、配列サイズが小さい場合、[-1,0,1,2、-1、-4,0,0,0,0,0,0,0,0,0,0,0,0 、0,0,0,0,0,0,0,0]の場合、実行には4ミリ秒かかりますが、アレイサイズが大きくなると、制限時間が超過するまで時間が増加し続けます。したがって、上記の解決策は制限時間を超えています。



以下はテストの入力ですが、上記のアルゴリズムは基本的に機能していません。

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上記の入力に対して、ブルートフォースアルゴリズムが実行され、結果は次のようになります。

Array length: 3000 1552391953319 [[0, 0, 0]] 1552392085789 Time-consuming: 132470 milliseconds

次に、主に配列要素の走査の時間計算量を減らすために、より良い解決策を見つける必要があります。解決すべき主な問題は、要素の3つのグループすべての配列をすばやく走査することです。



  • 配列配列
  • 3セットの要素をすばやくトラバースします
  • 繰り返す方法、完全な配置を取り除く

ネチズンによって与えられたより良い解決策を参照してください:

原則は、最初に配列を最初にソートすることです。O(n l o g 2 n)O(nlog_2 ^ n)、つまり、Arrays.sort(nums)です。外側のトラバーサルは、配列の先頭にあるポイント要素を固定し、内側のレイヤーは中央の要素を容易にするため、O(n 2)O(n ^ 2)上記のブルートフォースアルゴリズムよりも時間の再現性O(n 3)O(n ^ 3)nの効率を改善しました。

コードコメントの説明に重要なステートメントを作成しました。元の作成者の意図を理解しやすくするために、基本的な考え方は、左側の最初の要素を固定し、2番目の要素と3番目の要素に両端からアプローチし、要素を合理的に削除することです。アプローチプロセスで強調されているように、二重カウントを回避することも効率を改善するための鍵です。

他の人が敬意を払うためにそのようなコードを書くために!

// Position the first element, followed by the second element left, at the end an element right, left and right approaching the middle public List<List<Integer>> threeSumQuick(int[] nums) { ArrayList<List<Integer>> result = new ArrayList<List<Integer>>() if (nums.length < 3) return result Arrays.sort(nums) ArrayList<Integer> temp = null for (int i = 0 i < nums.length i++) { //Select nums[i] as the first number and de-weight it. Because the current element is equal to the previous element, there is no need to recalculate it once. if (i > 0 && nums[i] == nums[i - 1]) continue int left = i + 1 int right = nums.length - 1 while (right > left) { int sum = nums[i] + nums[left] + nums[right] if (sum == 0) { temp = new ArrayList<Integer>() temp.add(nums[i]) temp.add(nums[left]) temp.add(nums[right]) result.add(temp) // The left cursor is smaller than the right, and the left element is equal to the next element, then it is skipped, because it has been calculated once, and the weight is removed. while (left < right && nums[left] == nums[left + 1]) left++ // The left cursor +1 is smaller than the right, and the left element is equal to the previous element, then it is skipped, because it has been calculated once, and the weight is removed. while (left + 1 < right && nums[right] == nums[right - 1]) right-- } // and less than 0 means that the coordinate cursor needs to move right. The entire array is sorted in ascending order if (sum <= 0) left++ else if (sum >= 0) right-- } } return result }

上記の入力に対して、最適化アルゴリズムが実行され、結果は次のようになります。

Array length: 3000 1552392359849 [[0, 0, 0]] 1552392359858 Time-consuming: 9 milliseconds

130,000ミリ秒よりも9ミリ秒効率の向上は、2つのポイントではありません。他の人のアルゴリズムの利点を分析する必要があります。

楽しく遊ぶためには、アルゴリズムに関連する知識を補足する必要があります(王暁華のオンライン推奨 「アルゴリズムの楽しさ」 それを見ることができます)。より良い解決策を楽しみにしています。 。 。

参照

1、 3サム| Javaの最短コード実装
二、 マークダウン式編集研究ノート
3、 15.3Sum
4、 複雑さ、原理、実装など、一般的に使用される8つのソートアルゴリズムの詳細な分析
5、 バックトラッキングに関連する順列の組み合わせトラバーサル