3Sum

最新推荐文章于 2022-01-27 22:44:32 发布

pkufergus

最新推荐文章于 2022-01-27 22:44:32 发布

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分类专栏： leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/pkufergus/article/details/18357513

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本文介绍了一种解决三数之和问题的C++实现方案。通过对输入整数数组进行排序，并利用哈希表记录正负数值出现的次数，该算法能够有效地找到所有唯一三元组，使得这三个数相加为0。文章详细展示了如何遍历数组并利用哈希表快速检查所需配对是否存在。

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class Solution {
	public:
		vector<vector<int> > threeSum(vector<int> &num) {
			// Start typing your C/C++ solution below
			// DO NOT write int main() function
			//map< int, vector<int> > hashmap;
			map< int, int > hashneg;
			map< int, int > hashpos;
			vector< vector<int> > res;
			//map< vector<int>, int > hashres;
			vector<int>::iterator it;
			int i = 0;
			int c_zero = 0;
			int len = num.size();
			if (len < 3) {
			   		return res;
			}
			sort(num.begin(), num.end());
			vector<int> vec(len);
			i = 0;
			for (it = num.begin(); it != num.end(); it++) {
				//cout<<*it<<endl;
				vec[i] = *it;
				i++;
			}
			
			//return res;
			int zero_pos = 0;
			int pos = -1;
			int neg;
			int k,j, t;
			for (k = 0; k < len; k++) {
				if (vec[k] == 0) {
				   		zero_pos = k;
				   		c_zero++;
				} else if (vec[k] < 0) {
				  	   if (!hashneg.count(vec[k])) {
					   	  hashneg[vec[k]] = 1;
				  	   } else {
					   	 	  hashneg[vec[k]]++;
				  	   }
				} else {
				  	   if (!hashpos.count(vec[k])) {
					   	  hashpos[vec[k]] = 1;
				  	   } else {
					   	 	  hashpos[vec[k]]++;
				  	   }
				}
				
			}
			if (c_zero > 2) {
			   		   res.push_back(vector<int>());
			   		   res.back().push_back(0);
			   		   res.back().push_back(0);
			   		   res.back().push_back(0);
			}
			
			int a, b;
			int last = vec[0];
			int last_k = 10000;
			for (k = 0; vec[k] < 0; k++) {
				last = vec[0];
				if (vec[k] == last_k) {
				   		   continue;
				}
				//cout<<"neg k "<<vec[k]<<endl;
				for (j = k + 1; vec[j] < 0; j++) {
					if (vec[j] != last) {
					   		   //vec[j] = last;
					} else {
					  	   continue;
					}
				
					//cout<<"neg j "<<vec[j]<<endl;
					b = -(vec[k] + vec[j]);
					if (hashpos.count(b)) {
			  		   res.push_back(vector<int>());
			   		   res.back().push_back(vec[k]);
			   		   res.back().push_back(vec[j]);
			   		   res.back().push_back(b);
					}
					last = vec[j];
				}
				last_k = vec[k];
			}
			if (c_zero > 0) {
			   		   last = -1000;
   			   for (k = 0; vec[k] < 0; k++) {
			   	   if (vec[k] == last) {
				   	  		  continue;
			   	   }
					b = -(vec[k]);
					if (hashpos.count(b)) {
				  		   res.push_back(vector<int>());
				   		   res.back().push_back(vec[k]);
				   		   res.back().push_back(0);
				   		   res.back().push_back(b);
					}
					last = vec[k];
				}
			}
			last_k = -10000;
			for (k = 0; k < len - 1; k++) {
				last = -10000;
				if (vec[k] <= 0) {
				   		   continue;
				}
				if (vec[k] == last_k) {
				   		   continue;
				}
				//cout<<"pos"<<vec[k]<<endl;
				for (j = k + 1; j < len; j++) {
					if (vec[j] != last) {
					   		   //vec[j] = last;
					} else {
					  	   continue;
					}
					b = -(vec[k] + vec[j]);
					if (hashneg.count(b)) {
			  		   res.push_back(vector<int>());
			  		   res.back().push_back(b);
			   		   res.back().push_back(vec[k]);
			   		   res.back().push_back(vec[j]);
					}
					last = vec[j];
				}
				last_k =vec[k];
			}
			return res;
		}
};