[LeetCode] 3Sum

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本文介绍了一种算法，用于在一个整数数组中查找所有和为零的唯一三元组。该算法首先对数组进行排序，然后通过双指针技巧来有效地筛选出符合条件的三元组。

Total Accepted: 12950 Total Submissions: 78829

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 class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
        if (num.length < 3) return list;
        Arrays.sort(num);
        int sum = num[0] + num[1] + num[2];
        int closest = sum;
        
        for (int i = 0; i < num.length - 2; i++) {
            // skip replicated numbers
            if (i > 0 && num[i - 1] == num[i]) continue;
            if (num[i] > 0) break;
            
            int lo = i + 1;
            int hi = num.length - 1;
            
            while (lo < hi) {
                // skip replicated numbers
                while (lo < num.length - 1 && lo > i + 1 && num[lo] == num[lo-1]) lo++;
                while (hi < num.length - 1 && hi > i + 1 && num[hi] == num[hi+1]) hi--;
                if (lo >= hi) break;
                
                sum = num[i] + num[lo] + num[hi];  
                closest = Math.abs(sum) < Math.abs(closest) ? sum : closest;  
                
                if      (sum > 0)   hi--;  
                else if (sum < 0)   lo++;  
                else {
                    ArrayList<Integer> triplet = new ArrayList<Integer>();
                    triplet.add(num[i]);
                    triplet.add(num[lo]);
                    triplet.add(num[hi]);
                    
                    list.add(triplet);
                    hi--;
                    lo++;
                }
            }
        }
        
        return list;
    }
}