1.Two Sum (Array; HashTable)

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

法I：先快速排序，时间复杂度O(nlogn)，然后再二分法查找，时间发杂度O(nlogn)。这样比乱序时查找O(n2)要快。

struct MyStruct {
    int data;
    int pos;
    MyStruct(int d, int p){
        data = d;
        pos = p;
    }
    bool operator < (const MyStruct& rf) const{//参数用引用：避免占用过大内存，并且避免拷贝；用const防止改变操作数
        if(data < rf.data) return true;
        else return false;
    }
};
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<MyStruct> myNums;
        vector<int> ret;
        int idx1 = 0, idx2;
        for(int i = 0; i < nums.size(); i++){
            MyStruct* num = new MyStruct(nums[i],i);
            myNums.push_back(*num);
        }
        
        sort(myNums.begin(),myNums.end());
        while(!binarySearch(myNums, target-myNums[idx1].data, idx1+1, nums.size()-1, idx2)){
            idx1++;
        }
        
        ret.clear();
        if(myNums[idx1].pos < idx2){
            ret.push_back(myNums[idx1].pos+1);
            ret.push_back(idx2+1);
        }
        else{
            ret.push_back(idx2+1);
            ret.push_back(myNums[idx1].pos+1);
        }
        return ret; 
    }
    
    bool binarySearch(const vector<MyStruct>& nums, const int& target, int start, int end, int& targetPos)
    {
        if(end - start == 1){
            if(nums[start].data == target){
                targetPos = nums[start].pos;
                return true;
            }
            else if(nums[end].data == target){
                targetPos = nums[end].pos;
                return true;
            }
            else return false;
        }
        
        int mid = (start + end) >> 1;
        if(nums[mid].data == target){
            targetPos = nums[mid].pos;
            return true;
        } 
        
        if(nums[mid].data > target && start != mid && binarySearch(nums, target, start, mid, targetPos)) return true;
        if(binarySearch(nums, target, mid, end, targetPos)) return true;
        
        return false;
    }
};

法II：hash table. unordered_map的内部实现是先将key通过hash_value()得到其hash值，相同hash值的key以红黑树排列，查找时间是O(logn)，建树的时间类似建大顶堆，时间复杂度是O(n)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> ret;
        unordered_map<int,int> map;
        unordered_map<int,int>::iterator it;
        
        for(int i = 0; i < nums.size(); i++){
            it = map.find(target-nums[i]);
            if(it!=map.end()){
                ret.push_back(it->second);
                ret.push_back(i);
                return ret;
            }
            
            map.insert(make_pair(nums[i],i));
        }
        return ret;
    }
};