package amazon;


// 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: 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]
// ]

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * Created by anduo on 17-3-13.
 */
public class ThreeSum {

    // 把3sum的问题，降级到2sum问题
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums);

        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {// 这道题用不到这个条件，因为题目说了不考虑重复元素
                continue;
            }
            int left = i + 1;
            int right = nums.length - 1;
            int target = -nums[i];
            twoSum(nums, left, right, target, result);
        }
        return result;
    }

    public void twoSum(int[] nums,
                       int left,
                       int right,
                       int target,
                       List<List<Integer>> results) {
        // 充分利用有序这样一个特点
        while (left < right) {
            if (nums[left] + nums[right] == target) {
                ArrayList<Integer> triple = new ArrayList<>();
                triple.add(-target);
                triple.add(nums[left]);
                triple.add(nums[right]);
                results.add(triple);

                left++;
                right--;
                // skip duplicate pairs with the same left
                while (left < right && nums[left] == nums[left - 1]) {
                    left++;
                }
                // skip duplicate pairs with the same right
                while (left < right && nums[right] == nums[right + 1]) {
                    right--;
                }
            } else if (nums[left] + nums[right] < target) {
                left++;
            } else {
                right--;
            }
        }
    }

}