LeetCode刷题笔录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)

做过了2 Sum,这题自然不在话下。顺序遍历一遍数组，对于每个元素num[i],用-num[i]作为目标值对剩下的元素来做2 sum即可。注意数组排序以后，对于每一个i,只需要对i后面的数组元素进行2 sum即可。

另外这里比较trick的地方就是要注意去除重复元素。

public class Solution {
	public List<List<Integer>> threeSum(int[] num) {
		List<List<Integer>> res = new ArrayList<List<Integer>>();
		if (num.length <= 2)
			return res;
		Arrays.sort(num);

		int iValue = num[0];
		int jValue, kValue;
		List<Integer> sol = new ArrayList<Integer>(3);
		//initialize the arraylist with three meaningless values 
		for(int i = 0; i < 3; i++){
			sol.add(Integer.MIN_VALUE);
		}
		for (int i = 0; i < num.length - 2; i++) {
			// avoid duplication
			if (i != 0 && num[i] == iValue)
				continue;

			sol.set(0, num[i]);
			// do two sum
			int j = i + 1;
			int k = num.length - 1;

			while (j < k) {
				jValue = num[j];
				kValue = num[k];
				int twoSum = num[j] + num[k];
				// found a solution
				if (twoSum + sol.get(0) == 0) {

					sol.set(1, num[j]);
					sol.set(2, num[k]);

					res.add(new ArrayList<Integer>(sol));
					// increment j until num[j] is different from the current j
					// value to avoid duplicates
					while (++j < k && num[j] == jValue)
						;
					// the same for k
					while (--k > j && num[k] == kValue)
						;
				} else {
					if (twoSum + sol.get(0) < 0) {
						while (++j < k && num[j] == jValue)
							;
					} else {
						while (--k > j && num[k] == kValue)
							;
					}
				}
			}
			iValue = num[i];
		}

		return res;
	}
}