leetcode-15-three-sum

Description

Given an array nums of n integers, are there elements a, b, c in nums 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.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

####

/**
 * @note leetcode-三数之和为0
 * @apiNote   思路为 先排序， 然后有两个指针 head和end。 要计算的第一个数一定是负数,所以只要后两个数相加等于
 *  0-第一个数 即可。
 * @since 19-8-26 10:27 by jdk 1.8
 */
public class ThreeSum {

    private static List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        //进行排序
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) {
                return result;
            }
            //之前的相等，我选择跳过
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int header = i + 1;
            int end = nums.length - 1;
            int val = 0 - nums[i];
            while (header < end) {
                if (nums[header] + nums[end] == val) {
                    List<Integer> list = Arrays.asList(nums[i], nums[header], nums[end]);
                    result.add(list);
                    while (header < end && nums[end] == nums[end - 1]) end--;
                    while (header < end && nums[header] == nums[header + 1]) header++;
                    end--;
                    header++;
                } else if (nums[header] + nums[end] > val) {
                    end--;
                } else {
                    header++;
                }
            }
        }


        return result;
    }

    public static void main(String[] args) {
        int[] nums = {};
        List<List<Integer>> lists = threeSum(nums);
        System.out.println(lists);

    }
}

####

# O(n * n)
class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        nums.sort() # 排序
        for i in xrange(0, len(nums) - 2):
            # 如果与前一个值相同，则跳过
            # 避免重复计算
            if i > 0 and nums[i] == nums[i-1]:
                continue
            l, r = i + 1, len(nums) - 1
            while l < r:
                s = nums[i] + nums[l] + nums[r]
                if s < 0:
                    # 和过小，增大 nums[l]
                    l += 1
                elif s > 0:
                    # 值过大，减小 nums[r]
                    r -= 1
                else:
                    # 存储目标结果
                    res.append((nums[i], nums[l], nums[r]))
                    # 避免 l,r 重复计算
                    while l < r and nums[l] == nums[l + 1]:
                        l += 1
                    while l < r and nums[r] == nums[r - 1]:
                        r -= 1
                    # 下一组目标值
                    l += 1
                    r -= 1
        return res