[leetcode]Two Sum C++ python實現

方法一：用HASH表，將每個數x依次放入hash表中；然後檢查hash表有沒有有沒有target-x的數，有就輸出。

python實現：

class Solution(object):
    """docstring for ClassName"""
    def twoSum(self, nums, target):
        dict={}
        for i in xrange(len(nums)):
            x=nums[i]
            if target-x in dict:
                return (dict[target-x]+1, i+1)
            dict[x]=i

C++實現：

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
		map<int,int>mp;
		for(auto i=0;i<numbers.size();i++){
			mp[numbers[i]]=i;	
		}
		for(auto i=0;i<numbers.size();i++){
		auto j=mp.find(target-numbers[i]);
		
		if(j!=mp.end()){
		if(j->second==i)continue;
		vector<int>out;
			int k=j->second;
				out.push_back(k>i?i+1:k+1);
				out.push_back(k<i?i+1:k+1);
				return out;
			}			
		}
    }
};

方法二：先排序，記錄排序前的順序；然後用類似二分查找的辦法，找到相應的那個數。複雜度應該是O(nlogn)，注意在找target-x的時候從target/2開始查找可以節約一半的時間。

class Solution {
	vector<int>res;
	
public:
	vector<int> twoSum(vector<int>& nums, int target) {
		int len=nums.size();int right;
		vector<int>num2(nums.begin(),nums.end());
		sort(num2.begin(),num2.end());
		int Mid=Mydisearch(num2,target/2,0,len);
		for(int i=0;i<=Mid;i++){
			right=Mydisearch(num2,target-num2[i],Mid,len);
			if(num2[right]+num2[i]==target){
				int j=0;
				while(nums[j]!=num2[i])
					j++;
				res.push_back(j+1);
				int k=0;
				while(nums[k]!=num2[right])
					k++;
				if (k==j)
				{
					k++;
				}
				while(nums[k]!=num2[right])
					k++;
				res.push_back(k+1);
				sort(res.begin(),res.end());
				return res;
			}
		}
		return res;
	}
	int Mydisearch(vector<int>& nums, int target, int p, int q){
		int m = (p + q) /2;

		if(nums[m] == target|| m == p)
			return m;
		else if(nums[m]>target)
			return Mydisearch(nums,target,p,m);
		else
			return Mydisearch(nums,target,m+1,q);

	}
};