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ThreeSum.java
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ThreeSum.java
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package two_pointers;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**
* Created by gouthamvidyapradhan on 29/03/2017. 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.
*
* <p>Note: The solution set must not contain duplicate triplets.
*
* <p>For example, given array S = [-1, 0, 1, 2, -1, -4],
*
* <p>A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
*/
public class ThreeSum {
/**
* Main method
*
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception {
int[] nums = {
-1, 0, 1, 2, -1, -4, -1, 0, 1, 2, -1, -4, -1, 0, 1, 2, -1, -4, -1, 0, 1, 2, -1, -4, -1, 0, 1,
2, -1, -4, -1, 0, 1, 2, -1, -4, -1, 0, 1, 2, -1, -4, -1, 0, 1, 2, -1, -4
};
System.out.println(new ThreeSum().threeSum(nums));
}
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums.length < 3) return result;
Arrays.sort(nums);
for (int i = 0, l = nums.length; i < l - 2; i++) {
if (i == 0 || nums[i] != nums[i - 1]) {
int j = i + 1, k = l - 1;
while (k > j) {
if (j != i + 1 && nums[j] == nums[j - 1]) {
j++;
continue;
}
int sum = nums[i] + nums[j] + nums[k];
if (sum == 0) {
result.add(Arrays.asList(nums[i], nums[j], nums[k]));
k--;
j++;
} else if (sum > 0) k--;
else j++;
}
}
}
return result;
}
}