CODE 116. 3Sum

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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)

public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
	// IMPORTANT: Please reset any member data you declared, as
	// the same Solution instance will be reused for each test case.
	if (num.length <= 2) {
		ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
		return results;
	}
	ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
	Arrays.sort(num);
	for (int i = 0; i < num.length - 2; i++) {
		int sum = num[i];
		int start = i + 1;
		int end = num.length - 1;
		while (start < end) {
			if (num[start] + num[end] > 0 - sum) {
				while (start < end && num[end - 1] == num[end]) {
					end--;
				}
				end--;
			} else if (num[start] + num[end] < 0 - sum) {
				while (start < end && num[start + 1] == num[start]) {
					start++;
				}
				start++;
			} else {
				ArrayList<Integer> result = new ArrayList<Integer>();
				result.add(num[i]);
				result.add(num[start]);
				result.add(num[end]);
				if (!results.contains(result)) {
					results.add(result);
				}
				start++;
				end--;
			}
		}
		while (i < num.length - 2 && num[i + 1] == num[i]) {
			i++;
		}
	}
	return results;
}