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)

 Idea:  o(n^2) solution exists. First sort the array, and then from left to right, for each num[i], search the pair that sums up to -num[i] using Two Sum algorithm. 

 1 public class Solution {
 2     public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
 3         // Note: The Solution object is instantiated only once and is reused by each test case.
 4         Arrays.sort(num);
 5         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
 6         if(num == null || num.length < 3) return result;
 7         int len = num.length;
 8         for(int i = 0; i < num.length-2; i ++){
 9             if(i==0 || num[i]>num[i-1]){
10                 int j = i + 1;
11                 int h = len - 1;
12                 while(j < h){
13                     int sum = 0 - num[j] - num[h];
14                     if(sum == num[i]){
15                         ArrayList<Integer> row = new ArrayList<Integer>();
16                         row.add(num[i]);
17                         row.add(num[j]);
18                         row.add(num[h]);
19                         result.add(row);
20                         h--;
21                         j++;
22                         while(h>j && num[h]==num[h+1]) h--; 
23 
24                         while(j<h && num[j]==num[j-1]) j++;
25                     }else if(sum < num[i]){
26                             h --;
27                     }else if(sum > num[i]){
28                             j ++;
29                     }
30                 }
31             }
32         }
33         return result;
34     }
35 }

 

 第二遍：

 1 public class Solution {
 2     public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
 3         // Note: The Solution object is instantiated only once and is reused by each test case.
 4         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
 5         if(num == null || num.length < 3) return result;
 6         Arrays.sort(num);
 7         for(int i = 0; i < num.length - 2; i ++){
 8             if(i == 0 || num[i] > num[i-1]){
 9                 int target = - num[i];
10                 int j = i + 1;
11                 int h = num.length - 1;
12                 while(j < h){
13                     if(num[j] + num[h] == target){
14                         ArrayList<Integer> row = new ArrayList<Integer>();
15                         row.add(num[i]);
16                         row.add(num[j]);
17                         row.add(num[h]);
18                         result.add(row);
19                         int jnum = num[j];
20                         int hnum = num[h];
21                         while(num[j] == jnum && j < h) j ++;
22                         while(num[h] == hnum && h > j) h --;
23                     }else if(num[j] + num[h] > target){
24                         h --;
25                     }else{
26                         j ++;
27                     }
28                 }
29             }
30         }
31         return result;
32     }
33 }

 

posted on 2013-10-16 13:45  Step-BY-Step  阅读(148)  评论(0)  编辑  收藏  举报

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