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, 0, 0, 0, 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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 。