lintcode-medium-3 Sum

標籤：

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.

For example, given array S = {-1 0 1 2 -1 -4}, A solution set is:

(-1, 0, 1)(-1, -1, 2)

思路：先排序，然後逐個作為起點，再把後面的數從左右兩邊開始往中間搜尋，如果和為0就加入結果，如果大於零，right向左移動，如果小於零，left向右移動。注意去掉重複的情況。

public class Solution {    /**     * @param numbers : Give an array numbers of n integer     * @return : Find all unique triplets in the array which gives the sum of zero.     */    public ArrayList<ArrayList<Integer>> threeSum(int[] numbers) {        // write your code here        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();                if(numbers == null || numbers.length == 0)            return result;                Arrays.sort(numbers);                for(int i = 0; i < numbers.length - 2; i++){            if(i > 0 && numbers[i] == numbers[i - 1])                continue;                        int left = i + 1;            int right = numbers.length - 1;                        while(left < right){                if(numbers[i] + numbers[left] + numbers[right] == 0){                    ArrayList<Integer> line = new ArrayList<Integer>();                    line.add(numbers[i]);                    line.add(numbers[left]);                    line.add(numbers[right]);                    result.add(new ArrayList<Integer>(line));                                        left++;                    right--;                    while(left < right && numbers[left] == numbers[left - 1])                        left++;                    while(left < right && numbers[right] == numbers[right + 1])                        right--;                }                else if(numbers[i] + numbers[left] + numbers[right] < 0){                    left++;                }                else{                    right--;                }            }        }                return result;    }}

 

lintcode-medium-3 Sum