lintcode-medium-3 Sum

Posted 2020-06-21 哥布林工程师

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，然后逐个作为起点，再把后面的数从左右两边开始往中间搜索，如果和为0就加入结果，如果大于零，right向左移动，如果小于零，left向右移动。注意去掉重复的情况。

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;
    }
}