标签：des style blog color io os ar for sp
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)
难度：84， 我一开始采用brute force的方法，也就是NP做法，类似Combination Sum, Path Sum之类枚举的方法，结果毫不留情地在big case的时候TLE了，毕竟这里brute force复杂度O(N^3)。寻思无奈，看了一下别人的思路，Code Ganker提供了一种方法，先sort array, 然后3个元素中的一个元素num[i]从大到小遍历，另外两个元素用-num[i]作 target用Two Sum来找对应的集合。这里Two Sum 肯定不用brute force, 那样这里就O(N^2)了，也不用HashMap, 比较复杂。用夹逼方法来做，O(N)，当找到一个加起来为target的组合后，还不要忘了要skip一下重复的元素来节省时间。总的时间复杂度为O(NlongN + N^2) = O(N^2)
1 public class Solution { 2 public ArrayList<ArrayList<Integer>> threeSum(int[] num) { 3 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>(); 4 if (num == null || num.length < 3) { 5 return res; 6 } 7 Arrays.sort(num); 8 for (int i=num.length-1; i>=2; i--) { 9 if (i<num.length-1 && num[i]==num[i+1]) continue; 10 ArrayList<ArrayList<Integer>> sumOfTwo = twoSum(num, 0, i-1, -num[i]); 11 for (int k=0; k<sumOfTwo.size(); k++) { 12 ArrayList<Integer> temp = sumOfTwo.get(k); 13 temp.add(num[i]); 14 } 15 res.addAll(sumOfTwo); 16 } 17 return res; 18 } 19 20 public ArrayList<ArrayList<Integer>> twoSum(int[] num, int l, int r, int target) { 21 ArrayList<ArrayList<Integer>> sets = new ArrayList<ArrayList<Integer>>(); 22 while (l < r) { 23 if (num[l] + num[r] == target) { 24 ArrayList<Integer> set = new ArrayList<Integer>(); 25 set.add(num[l]); 26 set.add(num[r]); 27 sets.add(new ArrayList<Integer>(set)); 28 l++; 29 r--; 30 while (l<r && num[l]==num[l-1]) { 31 l++; 32 } 33 while (l<r && num[r]==num[r+1]) { 34 r--; 35 } 36 } 37 else if (num[l] + num[r] > target) { 38 r--; 39 } 40 else { 41 l++; 42 } 43 } 44 return sets; 45 } 46 }
标签：des style blog color io os ar for sp
原文地址：http://www.cnblogs.com/EdwardLiu/p/4010951.html
