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44_和为S的两个数字
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44_和为S的两个数字
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#include <iostream>
#include <vector>
using namespace std;
/*
题目描述:
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,使得他们的和正好是S,如果有多对数字的和等于S,输出两个数的乘积最小的。
*/
class Solution
{
public:
//很笨的方法 O(N)
vector<int> FindNumbersWithSum(vector<int> array, int sum)
{
vector<int> ret;
if (array.empty())
{
return ret;
}
for (size_t i = 0; i < array.size(); ++i)
{
if (array[i] >= sum)
{
break;
}
int val1 = array[i];
for (size_t j = i; j < array.size(); ++j)
{
if (array[j] >= sum)
{
break;
}
int val2 = array[j];
if (val1 + val2 == sum)
{
ret.push_back(val1);
ret.push_back(val2);
return ret;
}
}
}
return ret;
}
//改进
//O(N)
vector<int> FindNumbersWithSumImprove(vector<int> array, int sum)
{
vector<int> ret;
int begin = 0;
int end = array.size() - 1;
while (begin < end)
{
if (array[begin] * array[end] == sum)
{
ret.push_back(array[begin]);
ret.push_back(array[end]);
break;
}
while (array[begin] * array[end] < sum)
{
++begin;
}
while (array[begin] * array[end] > sum)
{
--end;
}
}
return ret;
}
};
void Test()
{
vector<int> ivec= { 1, 2, 3, 4, 5, 6 ,7 ,8, 9, 10 };
vector<int> ret = Solution().FindNumbersWithSum(ivec, 11);
for (int &i : ret)
{
cout << i << " ";
}
cout << endl;
}
int main()
{
Test();
return 0;
}