Two Sum

題目

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

分析

1.數字順序並非有序。

2.index1 < index2 ，並且index1與index2起始於1。

3.存在唯一解。

方案

1.存儲value和index對value進行排序，使用Quick Sort, 時間複雜度爲O(nlogn)，從兩端向中間逼近，複雜度爲O(n),總複雜度爲O(nlogn)。

2.Hash,C++中使用關聯容器map，map爲紅黑樹，查找效率爲O(logn),另外需要便利數組中每個元素，時間複雜度爲O(n)，故總時間複雜度爲O(nlogn)。

結論

從以上分析來看，時間複雜度相同，使用排序後查找佔用48ms，使用map佔用96ms。

CODE

struct Node{
    int value;
    int index;
};

bool cmp(Node a, Node b) {
    return a.value < b.value;
}

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector<Node> ary;
        for (int i = 0; i < numbers.size(); ++i) {
            Node t;
            t.value = numbers[i];
            t.index = i + 1;
            ary.push_back(t);
        }

        sort(ary.begin(), ary.end(),cmp);

        vector<int> ans;
        for (int i = 0, j = ary.size()-1; i < j; ) {
            int sum = ary[i].value + ary[j].value;
            if (sum == target) {
                int index1 = ary[i].index;
                int index2 = ary[j].index;
                if(index1 < index2) {
                    ans.push_back(index1);
                    ans.push_back(index2);
                } else {
                    ans.push_back(index2);
                    ans.push_back(index1);
                }
                break;
            } else if(sum < target) {
                i++;
            } else {
                j--;
            }
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
       map<int, int> mymap;
       for (int i = 0; i < numbers.size(); ++i) {
           mymap[numbers[i]] = i+1;
       }

       vector<int> result;
       for (int i = 0; i < numbers.size(); ++i) {
           int a = numbers[i];
           int b = target - a;
           map<int, int>::iterator it = mymap.begin();
           it = mymap.find(b);
           if (it != mymap.end()) {
               // find it
               int index1 = i+1;
               int index2 = it->second;
               if (index1 < index2) {
                   result.push_back(index1);
                   result.push_back(index2);
               } else if(index1 > index2) {
                   result.push_back(index2);
                   result.push_back(index2);
               } else {
                   continue;
               }
               break;
           }
       }
       return result;
    }
};