[leetcode] 3Sum

最新推荐文章于 2022-01-27 22:44:32 发布

原创最新推荐文章于 2022-01-27 22:44:32 发布 · 394 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

222 篇文章

订阅专栏

本文探讨了LeetCode上经典问题“三数之和”的解决方案。该问题要求从整数数组中找出所有不重复的三个数，使得它们的和为零，并确保找到的三元组按非递减顺序排列。文章提供了详细的C++和Java代码实现，通过排序和双指针技巧高效解决问题。

From : https://leetcode.com/problems/3sum/

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:

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)

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

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        if(null == nums || nums.length < 3) {
            return new ArrayList<List<Integer>>();
        }
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        
        for(int i=0, lst=nums.length-1; i<lst && nums[i]<=0; ++i) {
            if(i>0 && nums[i]==nums[i-1]) continue;
            int j = i+1, k = lst;
            while(j < k) {
                if(j>i+1 && nums[j]==nums[j-1]) {
                    ++j;
                    continue;
                }
                if(k<lst && nums[k]==nums[k+1]) {
                    --k;
                    continue;
                }
                int sum = nums[i]+nums[j]+nums[k];
                if(sum == 0) {
                    List<Integer> cur = new ArrayList<Integer>();
                    cur.add(nums[i]);
                    cur.add(nums[j++]);
                    cur.add(nums[k--]);
                    ans.add(cur);
                } else if(sum < 0) {
                    ++j;
                } else {
                    --k;
                }
            }
        }
        
        return ans;
    }
}