[LeetCode] 15. 3Sum 三数之和

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]
]

解法：排序 + 双指针。先对数组排序，然后遍历排序后的数组, 当循环到nums[i]时，对后面的数使用双指针left和right分别指向第一个和最后一个数，如果3个数的和等于0，就找到一组添加到结果中；如果小于0，说明和小了，要往大的方向移动，left指针右移1位；如果和大于0，说明大了，right指针左移1位。再看3个数的和。由于题目要求不能有重复的答案，所以对于每个指针移动后都要看是否和前面的相等，如果相等则跳过。

Time: O(n^2)  Space: O(1)

Java:

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public List<List<Integer>> threeSum(int[] num) {     Arrays.sort(num);     List<List<Integer>> res = new LinkedList<>();     for (int i = 0; i < num.length-2; i++) {         if (i == 0 || (i > 0 && num[i] != num[i-1])) {             int lo = i+1, hi = num.length-1, sum = 0 - num[i];             while (lo < hi) {                 if (num[lo] + num[hi] == sum) {                     res.add(Arrays.asList(num[i], num[lo], num[hi]));                     while (lo < hi && num[lo] == num[lo+1]) lo++;                     while (lo < hi && num[hi] == num[hi-1]) hi--;                     lo++; hi--;                 } else if (num[lo] + num[hi] < sum) lo++;                 else hi--;            }         }     }     return res; }

Python:

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def threeSum(self, nums):     res = []     nums.sort()     for i in xrange(len(nums)-2):         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:                 l +=1             elif s > 0:                 r -= 1             else:                 res.append((nums[i], nums[l], nums[r]))                 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

Python:

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class Solution(object):     def threeSum(self, nums):         """         :type nums: List[int]         :rtype: List[List[int]]         """         nums, result, i = sorted(nums), [], 0         while i < len(nums) - 2:             if i == 0 or nums[i] != nums[i - 1]:                 j, k = i + 1, len(nums) - 1                 while j < k:                     if nums[i] + nums[j] + nums[k] < 0:                         j += 1                     elif nums[i] + nums[j] + nums[k] > 0:                         k -= 1                     else:                         result.append([nums[i], nums[j], nums[k]])                         j, k = j + 1, k - 1                         while j < k and nums[j] == nums[j - 1]:                             j += 1                         while j < k and nums[k] == nums[k + 1]:                             k -= 1             i += 1         return result

Python: wo

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class Solution(object):     def threeSum(self, nums):         """         :type nums: List[int]         :rtype: List[List[int]]         """         res = []         nums = sorted(nums)         n = len(nums)         i = 0         while i < n - 2:             j, k = i + 1, n - 1             while j < k:                                s = nums[i] + nums[j] + nums[k]                 if s > 0:                     k -= 1                 elif s < 0:                     j += 1                 else:                     res.append([nums[i], nums[j], nums[k]])                     j += 1                     k -= 1                     while nums[j-1] == nums[j] and j < k:                         j += 1                     while nums[k+1] == nums[k] and j < k :                         k -= 1                                            i += 1             while nums[i-1] == nums[i] and i < n - 2:                 i += 1                       return res

C++：

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class Solution { public:     vector<vector<int>> threeSum(vector<int>& nums) {         vector<vector<int>> res;         sort(nums.begin(), nums.end());         for (int k = 0; k < nums.size(); ++k) {             if (nums[k] > 0) break;             if (k > 0 && nums[k] == nums[k - 1]) continue;             int target = 0 - nums[k];             int i = k + 1, j = nums.size() - 1;             while (i < j) {                 if (nums[i] + nums[j] == target) {                     res.push_back({nums[k], nums[i], nums[j]});                     while (i < j && nums[i] == nums[i + 1]) ++i;                     while (i < j && nums[j] == nums[j - 1]) --j;                     ++i; --j;                 } else if (nums[i] + nums[j] < target) ++i;                 else --j;             }         }         return res;     } };

 

类似题目：

[LeetCode] 16. 3Sum Closest 最近三数之和

[LeetCode] 259. 3Sum Smaller 三数之和较小值

 

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