15. 3Sum

题目描述和难度

  • 题目描述:

给定一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c ,使得 a + b + c = 0 ?找出所有满足条件且不重复的三元组。

注意:答案中不可以包含重复的三元组。

例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4],

满足要求的三元组集合为:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

思路分析

求解关键:

  • 我们可以借助“两数之和”的思路来解决“三数之和”,但是题目中要求我们去除重复,因此在我们可以先将数组元素排序。排序以后,我们对数组操作就比无序要方便得多。
  • 具体的技巧我们展示在两个参考解答中。

参考解答

参考解答1

Python 写法:

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        nums.sort()
        for i in range(len(nums) - 2):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            # 用指针对撞的方式
            l = i + 1
            r = len(nums) - 1
            # 不能等于,等于就变成取一样的数了
            while l < r:
                s = nums[i] + nums[l] + nums[r]
                if s > 0:
                    r -= 1
                elif s < 0:
                    l += 1
                else:
                    res.append([nums[i], nums[l], nums[r]])
                    # 注意:这一步在去重,是第一种解法 set 做不到的
                    while l < r and nums[l] == nums[l + 1]:
                        l += 1
                    while l < r and nums[r] == nums[r - 1]:
                        r -= 1
                    l += 1
                    r -= 1
        return res

if __name__ == '__main__':
    nums = [-2, 0, 0, 2, 2]
    solution = Solution()
    result = solution.threeSum(nums)
    print(result)

Java 写法:

import java.util.*;

public class Solution {

    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        int len = nums.length;
        if (len < 3) {
            return res;
        }
        Arrays.sort(nums);
        // 为了应对优化点 1 而对特殊测试用例的处理
        if (nums[0] == 0 && nums[len - 1] == 0) {
            List<Integer> threeSum = new ArrayList<>();
            threeSum.add(0);
            threeSum.add(0);
            threeSum.add(0);
            res.add(threeSum);
            return res;
        }

        // 注意分析边界条件
        // 1、最多只能到倒数第 3 位
        // 如果全部是 0 ,其实也符合题意,但是如果排序以后,第 1 个数都大于零,肯定不是解
        for (int i = 0; i < len - 2 && nums[i] <= 0; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            // target 一定 >= 0
            int target = -nums[i];
            int l = i + 1;
            int r = len - 1;
            while (l < r) {
                // 优化点1:如果右边的数都小于 0 了,左边的数也一定小于 0
                // 它们的和就不会大于 0
                if (nums[r] < 0) {
                    break;
                }
                int sum = nums[l] + nums[r];
                if (sum > target) {
                    r--;
                } else if (sum < target) {
                    l++;
                } else {
                    assert sum == target;
                    // 处理特例 int[] nums = {-2, 0, 0, 2, 2};
                    if (l > (i + 1) && nums[l] == nums[l - 1]) {
                        // 此时 nums[r] == nums[r + 1] 也一定成立
                        l++;
                        r--;
                        continue;
                    }
                    List<Integer> threeSum = new ArrayList<>();
                    threeSum.add(nums[i]);
                    threeSum.add(nums[l]);
                    threeSum.add(nums[r]);
                    res.add(threeSum);
                    l++;
                    r--;
                }
            }
        }
        return res;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] nums = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
        // int[] nums = {-1, 0, 1, 2, -1, -4};
        List<List<Integer>> threeSum = solution.threeSum(nums);
        System.out.println(threeSum);
    }
}

Java 写法:

import java.util.*;

public class Solution2 {

    // 时间复杂度 : O(n^2);
    // 空间复杂度 : O(n);

    /**
     * Edward 老师提供的解法
     */
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        int len = nums.length;
        if (len < 3) {
            return res;
        }
        Arrays.sort(nums);
        for (int i = 0; i < len - 2; i++) {
            // 避免 {-1, -1, -1, 1, 2, 3, 4} 这种情况出现
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int low = i + 1;
            int high = len - 1;
            int target = -nums[i];
            while (low < high) {
                int sum = nums[low] + nums[high];
                if (sum == target) {
                    res.add(Arrays.asList(nums[i], nums[low], nums[high]));
                    // 处理特例 int[] nums = {-2, 0, 0, 2, 2};
                    while (low < high && nums[low] == nums[low + 1]) {
                        low++;
                    }
                    while (low < high && nums[high] == nums[high - 1]) {
                        high--;
                    }
                    low++;
                    high--;
                } else if (sum < target) {
                    low++;
                } else {
                    high--;
                }
            }
        }
        return res;
    }

    public static void main(String[] args) {
        Solution2 solution2 = new Solution2();
        int[] nums = {-2, 0, 0, 2, 2};
        List<List<Integer>> threeSum = solution2.threeSum(nums);
        System.out.println(threeSum);
    }
}

参考解答

参考解答2:使用 two sum 的思路解决,要注意去重问题。


class Solution(object):
    # 排序可以去掉 -4 但是不能把后面重复的 2 去掉
    # [-4,-4,2,2]
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if len(nums) < 3:
            return []
        nums.sort()

        if nums[0] == nums[-1] == 0:
            return [[0, 0, 0]]

        res = set()
        # 最后两个数就没有必要作为遍历的起点了
        for index, one in enumerate(nums[:-2]):
            # 因为题目要求,答案中不可以包含重复的三元组。
            if index >= 1 and nums[index] == nums[index - 1]:
                continue
            s = set()
            for two in nums[index + 1:]:
                if two not in s:
                    s.add(-one - two)
                else:
                    # 找到了一个解
                    res.add((one, two, -one - two))
        return list(map(list, res))


if __name__ == '__main__':
    nums = [-2, 0, 0, 2, 2]
    solution = Solution()
    result = solution.threeSum(nums)
    print(result)

本篇文章的地址为 https://liweiwei1419.github.io/leetcode-solution/leetcode-0015-3sum ,如果我的题解有错误,或者您有更好的解法,欢迎您告诉我 liweiwei1419@gmail.com