[LeetCode1]3Sum

Given an array S of n integers, are there elements abc in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

  • Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
  • 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)
Analysis

Two-pointer scan

O(n^2)

c++

vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int>> result;
    sort(num.begin(),num.end());
    for(int i=0; i<num.size(); i++){
        int target = 0-num[i];
        int start = i+1;
        int end = num.size()-1;
        while(start<end){
            if(num[start]+ num[end] == target){
                vector<int> solution;
                solution.push_back(num[i]);
                solution.push_back(num[start]);
                solution.push_back(num[end]);
                result.push_back(solution);
                start++, end--;
                while(start<end && num[start] == num[start-1]) start++;
                while(start<end && num[end] == num[end+1]) end--;
            }
            else if(num[start] + num[end] >target)
                end--;
            else
                start++;
        }
        while(i<num.size() && num[i] == num[i+1]) i++;
    }
    return result;
    }
java

public List<List<Integer>> threeSum(int[] num) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
		Arrays.sort(num);
		int len = num.length;
		for(int i=0;i<len;i++){
			int target = 0-num[i];
			int start = i+1, end = len-1;
			while(start<end){
				if(num[start]+num[end]==target){
					ArrayList<Integer> content = new ArrayList<>();
    				content.add(num[i]);
    				content.add(num[start]);
    				content.add(num[end]);
    				result.add(content);
    				start++;
    				end--;
    				while(start<end &&num[start]==num[start-1]) start++;
    				while(start<end && num[end] == num[end+1]) end--;
				}
				else if(num[start]+num[end]>target){
					end--;
				}
				else {
					start++;
				}
			}
			
				while(i<len-1 && num[i]==num[i+1]) i++;
			
		}
		return result;
    }


在java中可以用Arrays的静态排序方法,。

用暴力搜索,三个循环,O(n^3)不能通过大数据

public List<List<Integer>> threeSum(int[] num) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        
        for(int i=0;i<num.length;i++){
        	int temp = 0-num[i];
        	for(int j=i+1;j<num.length;j++){
        		for(int k=j+1;k<num.length;k++){
        			if(num[j]+num[k]==temp){
        				ArrayList<Integer> content = new ArrayList<>();
        				content.add(i);
        				content.add(j);
        				content.add(k);
        				result.add(content);
        			}
        		}
        	}
        }
        return result;
    }



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