[LeetCode]15. 3Sum【&16. 3Sum Closest】
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15 . 3Sum
Medium
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]
]
回溯法:
41ms:
public List<List<Integer>> threeSum(int[] nums) { List<List<Integer>> res = new ArrayList<>(); if(nums==null||nums.length==0) return res; List<Integer> cur = new LinkedList<>(); Arrays.sort(nums); threesumdfs(res,cur,nums,0,0); return res;}private void threesumdfs(List<List<Integer>> res,List<Integer> cur,int[] nums, int sum,int start){ if(cur.size()==3){ if(sum==0) res.add(new ArrayList<Integer>(cur)); return ; } for(int i=start;i<nums.length;i++){ cur.add(nums[i]); threesumdfs(res,cur,nums,sum+nums[i],i+1); while(i<nums.length-1){ if(nums[i+1]==cur.get(cur.size()-1)) i++; else break; } cur.remove(cur.size()-1); }}
参考:
[LeetCode]39.Combination Sum&40.Combination Sum II&216.Combination Sum III&377.Combination Sum IV
两端夹逼:
public List<List<Integer>> threeSum(int[] nums){ List<List<Integer>> res = new ArrayList<>(); if(nums==null||nums.length<3) return res; Arrays.sort(nums); for(int i=0;i<nums.length-2;i++){ if(i>0&&nums[i]==nums[i-1]) continue; int j=i+1,k=nums.length-1; while(j<k){ if(nums[i]+nums[j]+nums[k]<0){ j++; while(j<k&&nums[j]==nums[j-1]) j++; }else if(nums[i]+nums[j]+nums[k]>0){ k--; while(j<k&&nums[k]==nums[k+1]) k--; } else{ List<Integer> cur = new ArrayList<>(); cur.add(nums[i]); cur.add(nums[j]); cur.add(nums[k]); res.add(cur); j++;k--; while(j<k&&nums[j]==nums[j-1]&&nums[k]==nums[k+1]) j++; } } } return res;}
参考:
1. Two Sum&167. Two Sum II - Input array is sorted
16 . 3Sum Closest
Medium
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
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