• 描述

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:
• Elementsinatriplet(a,b,c)mustbeinnon-descendingorder.(ie,a≤b≤c)
• Thesolutionsetmustnotcontainduplicatetriplets.

For example, given array S={-1 0 1 2 -1 4}

A solution set is:
(-1 0 1)
(-1 -1 2)

• 分析

有序Two Sum的衍生版，那么可以沿用Two Sum，使Three Sum变成Two Sum，在外层套一个循环即可，时间复杂度为O(n^2)

package leet.ThreeSum;

import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;


public class Solution {
    public static List<Integer> getThreeSum(int[] arr, int target) {
        List<Integer> res = new ArrayList<Integer>(6);
        if(arr == null || arr.length < 3)
            return res;
        // 为保证结果集有序，要先排序
        Arrays.sort(arr);
        for(int i = 0; i < arr.length-2; ++i) {
            int low = i + 1;
            int high = arr.length - 1;
            // 确保结果不会重复
            if(i>0 && arr[i] == arr[i-1]) continue;
            int sum = target - arr[i];
            // Two Sum的解决方法，不同的是要排除重复的，如果重复，则跳过
            while(low < high) {
                if(arr[low] + arr[high] > sum) {
                    -- high;
                    while(arr[high] == arr[high+1] && low < high)
                        -- high;
                }
                else if(arr[low] + arr[high] < sum) {
                    ++low;
                    while (arr[low] == arr[low - 1] && low < high)
                        ++low;
                }
                else {
                    res.add(arr[i]);
                    res.add(arr[low]);
                    res.add(arr[high]);
                    ++ low;
                    -- high;
                    while (arr[low] == arr[low - 1] && arr[high] == arr[high+1] && low < high)
                        ++low;
                }
            }
        }
        return res;

    }
}