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题目：
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)
思路：
1）用递归，先排序，确定第一个，然后确定第二个，再寻找第三个。
package sum; import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class ThreeSum { public List<List<Integer>> threeSum(int[] nums) { List<List<Integer>> res = new ArrayList<List<Integer>>(); int len; if (nums == null || (len = nums.length) < 3) return res; Arrays.sort(nums); for (int i = 0; i < len - 2; ++i) { int rem = 0 - nums[i]; List<Integer> subRes = new ArrayList<Integer>(); subRes.add(nums[i]); GetTwo(rem, i + 1, nums, len, subRes, res); // Move forward if next element is the same as current element while (i < len - 1 && nums[i+1] == nums[i]) ++i; } return res; } private void GetTwo(int rem, int start, int[] nums, int len, List<Integer> subRes, List<List<Integer>> res) { for (int i = start; i < len - 1; ++i) { int last = rem - nums[i]; List<Integer> cpySubRes = new ArrayList<Integer>(subRes); cpySubRes.add(nums[i]); GetLast(last, i + 1, nums, len, cpySubRes, res); // Move forward if next element is the same as current element while (i < len - 1 && nums[i+1] == nums[i]) ++i; } } private void GetLast(int rem, int start, int[] nums, int len, List<Integer> subRes, List<List<Integer>> res) { for (int i = start; i < len; ++i) { if (rem == nums[i]) { subRes.add(nums[i]); res.add(subRes); break; } } } public static void main(String[] args) { // TODO Auto-generated method stub ThreeSum t = new ThreeSum(); int[] S = { -1, 0, 1, 2, -1, -4 }; List<List<Integer>> res = t.threeSum(S); for (List<Integer> subRes : res) { for (int i : subRes) System.out.print(i + "\t"); System.out.println("\n"); } } }
2）非递归；排完序之后确立第一个元素，然后用两个指针指向剩下元素的头和尾，两边一夹，然后移动。
package sum; import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class ThreeSum { public List<List<Integer>> threeSum(int[] nums) { List<List<Integer>> res = new ArrayList<List<Integer>>(); int len; if (nums == null || (len = nums.length) < 3) return res; Arrays.sort(nums); for (int i = 0; i < len - 2;) { int rem = 0 - nums[i]; int left = i + 1; int right = len - 1; while (left < right) { if (nums[left] + nums[right] == rem) { List<Integer> subRes = new ArrayList<Integer>(); subRes.add(nums[i]); subRes.add(nums[left]); subRes.add(nums[right]); res.add(subRes); // Filter the duplicated elements. do { ++left; } while (left < len && nums[left] == nums[left - 1]); do { --right; } while (right >= 0 && nums[right + 1] == nums[right]); } else if (nums[left] + nums[right] < rem) { left++; } else { right--; } } // Move forward if the next element is the same as current element. do { ++i; } while (i < len && nums[i] == nums[i - 1]); } return res; } public static void main(String[] args) { // TODO Auto-generated method stub ThreeSum t = new ThreeSum(); int[] S = { -2,0,1,1,2 }; List<List<Integer>> res = t.threeSum(S); for (List<Integer> subRes : res) { for (int i : subRes) System.out.print(i + "\t"); System.out.println("\n"); } } }
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5025767.html
