【Leetcode】3Sum
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题目:
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]]思路:
暴力解决法是每个人都能想到的,三层for循环,时间复杂度是O(n^3),而且还要处理重复的问题,显然不是题目想要的解法。
我们不妨先对数组排个序。排序之后,我们就可以对数组用两个指针分别从前后两端向中间扫描了,如果是 2Sum,我们找到两个指针之和为target就OK了,那 3Sum 类似,我们可以先固定一个数,然后找另外两个数之和为第一个数的相反数就可以了。
Java版本:
public List<List<Integer>> threeSum(int[] nums) { List<List<Integer>> res = new ArrayList<>(); Arrays.sort(nums); for (int i = 0; i + 2 < nums.length; i++) { if (i > 0 && nums[i] == nums[i - 1]) { // skip same result continue; } int j = i + 1, k = nums.length - 1; int target = -nums[i]; while (j < k) { if (nums[j] + nums[k] == target) { res.add(Arrays.asList(nums[i], nums[j], nums[k])); j++; k--; while (j < k && nums[j] == nums[j - 1]) j++; // skip same result while (j < k && nums[k] == nums[k + 1]) k--; // skip same result } else if (nums[j] + nums[k] > target) { k--; } else { j++; } } } return res;}首先数组长度至少是3,固定一个数值索引i,并且保证紧接着的数不与前面的数值相同,若当前nums[i]与nums[i-1]相同,说明这个值已经判断过了,往后再移动一位取值,索引为j,从数组后面取值为k,目标值target为nums[i]的负数,由两边向中间靠拢,由于是从小到大的排序,当匹配到target后,j++,k--,否则,若大于target,则k--,即减大的数,若小于target,则j++,增大小的数,当j>k时,说明一次循环检查结束。
Python版本:
class Solution: 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 r = i+1 k = len(nums)-1 while r < k: s = nums[i] + nums[k] + nums[r] if s < 0: r +=1 elif s > 0: k -= 1 else: res.append((nums[i], nums[r], nums[k])) r += 1; k -= 1 while r < k and nums[r] == nums[r-1]: r += 1 while r < k and nums[k] == nums[k+1]: k -= 1 return res
C++版本:
vector<vector<int> > threeSum(vector<int> &num) { vector<vector<int> > res; std::sort(num.begin(), num.end()); for (int i = 0; i < num.size(); i++) { int target = -num[i]; int front = i + 1; int back = num.size() - 1; while (front < back) { int sum = num[front] + num[back]; // Finding answer which start from number num[i] if (sum < target) front++; else if (sum > target) back--; else { vector<int> triplet(3, 0); triplet[0] = num[i]; triplet[1] = num[front]; triplet[2] = num[back]; res.push_back(triplet); // Processing duplicates of Number 2 // Rolling the front pointer to the next different number forwards while (front < back && num[front] == triplet[1]) front++; // Processing duplicates of Number 3 // Rolling the back pointer to the next different number backwards while (front < back && num[back] == triplet[2]) rear--; } } // Processing duplicates of Number 1 while (i + 1 < num.size() && num[i + 1] == num[i]) i++; } return res; }
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