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)

public class Solution {
    private static class ThreeSumEntry {
        public int begin;
        public int mid;
        public int end;

        private ThreeSumEntry(int begin, int mid, int end) {
            this.begin = begin;
            this.mid = mid;
            this.end = end;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;

            ThreeSumEntry that = (ThreeSumEntry) o;

            if (begin != that.begin) return false;
            if (end != that.end) return false;
            if (mid != that.mid) return false;

            return true;
        }

        @Override
        public int hashCode() {
            int result = begin;
            result = 31 * result + mid;
            result = 31 * result + end;
            return result;
        }
    }
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (num==null || num.length < 3) {
            return result;
        }
        Arrays.sort(num);
        //用于结果的去重复
        HashMap<ThreeSumEntry,Boolean> map = new HashMap<ThreeSumEntry, Boolean>();
        //固定一个数，然后转换成对一个序列求和为指定数字的两个
        for (int i=0;i<num.length;i++) {
            int begin = 0,end = num.length-1;
            int find2Sum = 0-num[i];
            while (begin < end) {
                //是固定的数 就跳过
                if (begin==i) {
                    begin++;
                    continue;
                }
                if (end==i) {
                    end--;
                    continue;
                }
                if ((num[begin]+num[end])==find2Sum) {
                    ArrayList<Integer> sum = new ArrayList<Integer>();
                    int[] temp = {num[i],num[begin],num[end]};
                    Arrays.sort(temp);
                    ThreeSumEntry ts = new ThreeSumEntry(temp[0],temp[1],temp[2]);
                    if (map.containsKey(ts)) {
                        //如果是个重复的结果的数就跳过
                        begin++;
                        end--;
                        continue;
                    } else {
                        map.put(ts,true);
                    }
                    sum.add(temp[0]);
                    sum.add(temp[1]);
                    sum.add(temp[2]);
                    result.add(sum);
                    //找到一个结果还需要继续查找
                    begin++;
                    end--;
                    continue;
                }
                if ((num[begin]+num[end])>find2Sum) {
                    end--;
                }
                if ((num[begin]+num[end])<find2Sum) {
                    begin++;
                }
            }

        }
        return result;
    }
}