【中级算法】1.三数之和

题目:

给定一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c ,使得 a + b + c = 0 ?找出所有满足条件且不重复的三元组。

注意:答案中不可以包含重复的三元组。

例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4],

满足要求的三元组集合为:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

解题思路:

class Solution {
public:
/*
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        int len = nums.size();
        
        if(nums.size() < 3){
            return res;
        }
        
        for(int i = nums.size()-1;i >=2;--i){
            for(int j = i-1;j >= 1;--j){
                for(int k = j-1;k >=0;--k){
                    if(nums[i] + nums[j] + nums[k] == 0){
                        vector<int> tmp;
                        tmp.push_back(nums[i]);
                        tmp.push_back(nums[j]);
                        tmp.push_back(nums[k]);
                        res.push_back(tmp);
                    }
                }
            }
        }
        
        return res;
    }
*/
    int binarySearch(vector<int>& nums,int target,int start,int end){
        if(start > end){
            return -1;
        }
        
        while(start <= end){
            int mid = (start + end)/2;
            if(nums[mid] == target){
                return mid;
            }
            if(nums[mid] > target){
                end = mid - 1;
            }
            if(nums[mid] < target){
                start = mid + 1;
            }
        }
        
        return -1;
    }
    
    bool findTree(vector<vector<int>> nums,int a,int b,int c){
        for(int i = 0;i < nums.size(); ++i){
            if(nums[i][0] == a &&
               nums[i][1] == b &&
               nums[i][2] == c){
                return true;
            }
        }
        
        return false;
    }
    
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        int len = nums.size();
        int target = 0;
        if(len < 3){
            return res;
        }
        
        sort(nums.begin(),nums.end());
        
        for(int i = 0;i < len-2 ; ++i){
            if(i > 0 && nums[i-1] == nums[i]){
                continue;
            }
            for(int j = i + 1,k = len - 1 ;j < k;){
                if(j > i + 1 && nums[j] == nums[j-1]){
                    j++;
                    continue;
                }
                if(k < len -1&& nums[k] == nums[k+1]){
                    k--;
                    continue;
                }
                int sum = nums[i] + nums[j] + nums[k];
                if(sum > target){
                    k--;
                }else if(sum < target){
                    j++;
                }else{
                    vector<int> tmp;
                    tmp.push_back(nums[i]);
                    tmp.push_back(nums[j]);
                    tmp.push_back(nums[k]);
                    res.push_back(tmp);
                    j++;
                    k--;    
                }
            }
        }
        
        return res;
    }

};
原文地址:https://www.cnblogs.com/mikemeng/p/9011634.html