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　　　　leetcode 练习1  two sum

　　　　　　　　　　　　whowhoha@outlook.com

问题描述

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].
解法1：暴力破解法: O(n^2) runtime, O(1) space – Brute force: The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through

the rest of array, its runtime complexity is O(n^2).

解法2:使用HashMap。把每个数都存入map中，然后再逐个遍历，查找是否有 target – nums[i]。  O(n) runtime  O(n) space,

vector<int> twoSum(vector<int>& nums, int target){

       vector<int> vec;

       map<int,int> m;

       for (int i = 0; i < nums.size(); i++)

       {

              if(m.find(target - nums[i]) != m.end())

              {

                     vec.push_back(m[target - nums[i]] );

                     vec.push_back(i);

                     break;

              }

              m[nums[i]] = i;

       }

       return vec;

}

调用：

       int a[6]={2,7,1,8,9};

       vector<int> vec(a,a+5);

       vector <int> vect= twoSum(vec,15);

       return 1;

Two Sum II – Input array is sorted

Question:

Similar to Question [1. Two Sum], except that the input array is already sorted in

ascending order.

同上题：先排序，然后从开头和结尾同时向中间查找，原理也比较简单。O(nlogn) runtime, O(1) space

      vector<int> twoSumSored(vector<int>& nums, int target){

         vector<int> vecCopy(nums);

         int i=0,n=2,l=0,r = nums.size()-1;

         sort(vecCopy.begin(),vecCopy.end());

         int j=nums.size()-1;

         while(i<j)

         {

                   int sum = nums[i]+nums[j];

                   if (sum <target)

                   {

                            i++;

                   }

                   else if (sum > target)

                   {

                            j--;

                   }

                   else

                   {

                            vector<int> index;

                            index.push_back(i+1);

                            index.push_back(j+1);

                            return index;

                   }

         }

}

备注：vector 初始化方法，，map的使用方法

Two Sum III – Data structure design

Question:

Design and implement a TwoSum class. It should support the following operations: add and find.

add(input) – Add the number input to an internal data structure.

find(value) – Find if there exists any pair of numbers which sum is equal to the value.

For example,

add(1); add(3); add(5);

find(4) ? true;

 find(7) ? false

      我们把每个数字和其出现的次数建立映射，然后遍历哈希表，对于每个值，我们先求出此值和目标值之间的差值t，然后分两种情况来看，如果当前值不等于差值t，那么只要哈希表中有差值t就返回True，或者是当差值t等于当前值时，如果此时哈希表的映射次数大于1，则表示哈希表中还有另一个和当前值相等的数字，二者相加就是目标值。

class TwoSum { 

private: 

         unordered_map<int,int> table; 

public: 

         void add(int number) { 

                   table[number]++; 

         } 

         bool find(int value) { 

                   for(auto elem : table)

                   { 

                            for (vector<string>::iterator i = v.begin(); i != v.end(); ++i);

                            for (auto i = v.begin(); i != v.end(); ++i);

                            int num1 = elem.second; 

                            int num2 = value == (elem.first << 1) ? num1 - 1 : buffer.find(value - elem.first) != buffer.end(); 

                            if(num1 && num2){ 

                                     return true; 

                            } 

                   } 

                   return false; 

         } 

};  

leetcode 练习1 two sum

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原文地址：http://www.cnblogs.com/whowhoha/p/5738282.html