剑指offer 和为S的两个数字

输入一个递增排序的数组和一个数字S，在数组中查找两个数，是的他们的和正好是S，如果有多对数字的和等于S，输出两个数的乘积最小的。 

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该算法没有利用数组是有序的这个信息，虽然算法时间也是O(n)，但是空间消耗也是O(n)

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#include <stdafx.h> #include<vector> #include<iostream> #include <unordered_map> using namespace std; vector<int > FindNumbersWithSum( vector<int > array, int sum ) {                  vector<int > result;                  int size =array. size();                  if(size <2)return result;                  unordered_map<int ,int> m_map;                  for(int i=0; i<size ;i++){                                  m_map[array [i]]= i;                 }                  vector<vector <int> > results;                  for(int i=0; i<size ;i++){                                  const int gap= sum-array [i];                                  if(m_map .find( gap)!=m_map .end() && m_map[gap ]>i){                                                  result.clear ();                                                  result.push_back (array[ i]);                                                  result.push_back (gap);                                                  results.push_back (result);                                      }                 }                  int results_num =results. size();                  int i_index =0,time=999999999;                  if(results_num <2)return result;                  else                 {                                  for(int i=0; i<results_num ;i++){                                                  int tmp =results[ i][0]*results [i][1];                                                  if(tmp <time){                                                                  time=tmp ;                                                                  i_index=i ;                                                 }                                 }                                  result=results [i_index];                 }                  return result ; } int main () {                  int a [20]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};                  vector<int > test( a,a +20);                  vector<int >result= FindNumbersWithSum(test ,21);                  return 0; }

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时间消耗O(n)，空间消耗O(1)

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class Solution { public:     vector<int> FindNumbersWithSum(vector<int> array,int sum) {         int i = 0;         int j = array.size()-1;         vector<int> result;         while(i < j)         {             if(array[i] + array[j] == sum)             {                 result.push_back(array[i]);                 result.push_back(array[j]);                 return result;             }             else if(array[i] + array[j] > sum)                 j--;             else                 i++;         }         return result;     } };

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