3Sum
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)
最开始想着直接用穷举法，三重循环，时间复杂度O(N^3)略高，TLE超时，提示说用双指针，百度之，看到http://blog.csdn.net/xiaobaohe/article/details/7857461
用的是穷举法，不过优化了一下，先讲数组排序，然后去重，效率高了一点，600ms左右，java做的
后面百度看双指针如何求解
1 package com.gxf.test; 2 3 4 import java.util.ArrayList; 5 import java.util.Arrays; 6 import java.util.HashMap; 7 8 import java.util.List; 9 10 11 //class ListNode { 12 // public int val; 13 // public ListNode next; 14 // ListNode(int x) { 15 // val = x; 16 // next = null; 17 // } 18 // } 19 20 public class Solution { 21 public List<List<Integer>> threeSum(int[] num) { 22 List<List<Integer>> result = new ArrayList<List<Integer>>(); 23 HashMap<String, Integer> map = new HashMap<String, Integer>(); 24 Arrays.sort(num); 25 // for(int i = 0; i < num.length; i++){ 26 // System.out.print(num[i] + " "); 27 // } 28 // System.out.println(); 29 30 for(int i = 0; i < num.length - 2; i++){ 31 if(num[i] > 0) 32 break; 33 if(i > 0 && num[i] == num[i - 1]) 34 continue; //去重 35 for(int j = i + 1; j < num.length - 1; j++){ 36 if(j > i + 1 && num[j] == num[j - 1]) 37 continue; //去重 38 for(int k = j + 1; k < num.length; k++){ 39 if(k > j + 1 && num[k] == num[k - 1]) 40 continue; //去重 41 if(num[i] + num[j] + num[k] == 0){ 42 List<Integer> element = new ArrayList<Integer>(); 43 44 element.add(num[i]); 45 element.add(num[j]); 46 element.add(num[k]); 47 48 result.add(element); 49 } 50 } 51 } 52 } 53 54 55 return result; 56 } 57 }
下面是用双指针做的，时间复杂度好像为O(n*logn)
1 import java.util.ArrayList; 2 import java.util.Arrays; 3 import java.util.HashMap; 4 5 import java.util.List; 6 7 8 //class ListNode { 9 // public int val; 10 // public ListNode next; 11 // ListNode(int x) { 12 // val = x; 13 // next = null; 14 // } 15 // } 16 17 public class Solution { 18 public List<List<Integer>> threeSum(int[] num) { 19 List<List<Integer>> result = new ArrayList<List<Integer>>(); 20 Arrays.sort(num); //将数组升序排序 21 22 for(int i = 0; i < num.length; i++){ 23 if(i > 0 && num[i] == num[i - 1]) 24 continue; //去重 25 int j = i + 1; 26 int k = num.length - 1; 27 while(j < k){ //从i + 1 ~ n - 1中找出两个数等于-num[i] 28 if(j > i + 1 && num[j] == num[j - 1]) 29 { 30 j++; 31 continue; 32 } 33 if(k < num.length - 1 && num[k] == num[k + 1]){ 34 k--; 35 continue; 36 } //去重 37 int temp = num[i] + num[j] + num[k]; 38 if(temp > 0){ //结果比0大k前移 39 k--; 40 continue; 41 } 42 else if(temp < 0){ 43 j++; 44 continue; 45 }else{ //找到一个解 46 List<Integer> element = new ArrayList<Integer>(); 47 element.add(num[i]); 48 element.add(num[j]); 49 element.add(num[k]); 50 result.add(element); 51 j++; 52 //break; //退出循环 53 } 54 55 } 56 } 57 58 return result; 59 } 60 }
参考：http://www.cnblogs.com/lihaozy/archive/2013/01/23/2872427.html