1.Two Sum (Array; Divide-and-Conquer)

標籤：

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

思路：先快速排序，時間複雜度O(nlogn)，然後再二分法尋找，時間發雜度O(nlogn)。這樣比亂序時尋找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{//參數用引用：避免佔用過大記憶體，並且避免拷貝；用const防止改變運算元        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;    }};

 

1.Two Sum (Array; Divide-and-Conquer)