[LeetCode1]3Sum

本文主要是介绍[LeetCode1]3Sum，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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 ≤ 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, 0, 1)(-1, -1, 2)

Analysis

Two-pointer scan

O(n^2)

c++

vector<vector<int> > threeSum(vector<int> &num) {vector<vector<int>> result;sort(num.begin(),num.end());for(int i=0; i<num.size(); i++){int target = 0-num[i];int start = i+1;int end = num.size()-1;while(start<end){if(num[start]+ num[end] == target){vector<int> solution;solution.push_back(num[i]);solution.push_back(num[start]);solution.push_back(num[end]);result.push_back(solution);start++, end--;while(start<end && num[start] == num[start-1]) start++;while(start<end && num[end] == num[end+1]) end--;}else if(num[start] + num[end] >target)end--;elsestart++;}while(i<num.size() && num[i] == num[i+1]) i++;}return result;}

java

public List<List<Integer>> threeSum(int[] num) {List<List<Integer>> result = new ArrayList<List<Integer>>();Arrays.sort(num);int len = num.length;for(int i=0;i<len;i++){int target = 0-num[i];int start = i+1, end = len-1;while(start<end){if(num[start]+num[end]==target){ArrayList<Integer> content = new ArrayList<>();content.add(num[i]);content.add(num[start]);content.add(num[end]);result.add(content);start++;end--;while(start<end &&num[start]==num[start-1]) start++;while(start<end && num[end] == num[end+1]) end--;}else if(num[start]+num[end]>target){end--;}else {start++;}}while(i<len-1 && num[i]==num[i+1]) i++;}return result;}

在java中可以用Arrays的静态排序方法，。

用暴力搜索，三个循环，O(n^3)不能通过大数据

public List<List<Integer>> threeSum(int[] num) {List<List<Integer>> result = new ArrayList<List<Integer>>();for(int i=0;i<num.length;i++){int temp = 0-num[i];for(int j=i+1;j<num.length;j++){for(int k=j+1;k<num.length;k++){if(num[j]+num[k]==temp){ArrayList<Integer> content = new ArrayList<>();content.add(i);content.add(j);content.add(k);result.add(content);}}}}return result;}

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