(算法分析Week2)3Sum[Medium]
2017-09-15 21:55
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15. 3Sum[Medium]
Description
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.
Example
For example, given array S = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
Solution
刚开始第一个念头就是三重循环(太暴力了….)不过想想也不能过就没有尝试。做2Sum的时候就想过如果不是要返回对应位置就可以先排序然后再从前之后从后往前遍历一次。Two Sum这里要求返回具体的数字,所以可以用这个思路。
(1)排序,从小到大
(2)从下标为0开始遍历整个数组,target=0 - arr[i]
(3)分三种情况,等于target时将数组加入vector,小于target和大于target时则移动指针
(4)特殊考虑
①重复情况:因为不允许有重复的数组,所以在移动指针和取target的时候要先判断相邻两个数字是否相同,若相同则跳过
②及时跳出:当循环到arr[i]大于0时,说明已经结束,因为arr[i+1]≥ arr[i]。
Complexity analysis
for循环遍历一次算O(N),内层循环遍历N-i,总的来说是O(N²)Code
class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> result; int target = 0; int left, right; left = right = 0; sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size(); i++) { if (nums[i] > 0) break; //升序,如果已经大于0,说明后面的都大于0,不可能符合条件 if (nums[i] == nums[i-1] && i > 0) //这里要注意加上i>0的条件,否则越界【会RE】 continue; //排除重复 target = 0 - nums[i]; left = i+1; right = nums.size() - 1; while(left < right) { if (nums[left] + nums[right] == target) { result.push_back({nums[i], nums[left], nums[right]}); //继续排除重复 while(left < right && nums[left] == nums[left+1]) { left++; } while(left < right && nums[right] == nums[right-1]) { right--; } left++; right--; //完成一组,需要继续往下 } else if (nums[left] + nums[right] < target) { left++; } else if (nums[left] + nums[right] > target) { right--; } } } return result; } };
Result
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