Sum3.java ( File view )

  • By issaczr 2014-10-24
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  • Point(s): 1
			// Author:   Li Long, 1988lilong@163.com
// Date:     Apr 17, 2014
// Source:   http://oj.leetcode.com/problems/3sum/
// Analysis: http://blog.csdn.net/lilong_dream/article/details/21645437

// 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)

import java.util.ArrayList;
import java.util.Arrays;

public class Sum3 {

	public ArrayList<ArrayList<Integer>> threeSum(int[] num) {

		ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

		Arrays.sort(num);

		for (int i = 0; i < num.length - 2; ++i) {

			if (i > 0 && num[i] == num[i - 1]) {

				continue;
			
}
			
			int j = i + 1;
			int k = num.length - 1;

			while (j < k) {

				if (num[i] + num[j] + num[k] > 0) {

					--k;
				
} else if (num[i] + num[j] + num[k] < 0) {

					++j;
				
} else {

					ArrayList<Integer> tmp = new ArrayList<Integer>();
					tmp.add(num[i]);
					tmp.add(num[j]);
					tmp.add(num[k]);

					result.add(tmp);
					
					--k;
					++j;
					
					while (j < k && num[j] == num[j - 1]) {

						++j;
					
}
					while (j < k && num[k] == num[k + 1]) {

						--k;
					
}
				
}
			
}
		
}

		return result;
	
}

	public static void main(String[] args) {

		int[] num = {
 -1, 0, 1, 2, -1, -4 
};
		
		Sum3 slt = new Sum3();
		ArrayList<ArrayList<Integer>> result = slt.threeSum(num);

		for (ArrayList<Integer> array : result) {

			System.out.println(array);
		
}
	
}

}
			
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