來自劍指offer

#include "stdafx.h"
#include <iostream>
using namespace std;

bool FindNumbersWithSum(int* data, int length, int sum, int* num1, int* num2)
{
	bool bFound = false;
	if(data==NULL || length < 1)
		return bFound;

	int ahead = length-1;
	int behind = 0;

	while ( ahead > behind )
	{
		int curSum = data[ahead] + data[behind];
		if (curSum == sum)
		{
			*num1 = data[behind];
			*num2 = data[ahead];
			bFound = true;
			break;
		}
		else if (curSum > sum)
			ahead--;
		else
			behind++;
	}

	return bFound;
}

int _tmain(int argc, _TCHAR* argv[])
{
	int a[] = {1,2,4,7,11,15};
	int num1, num2;
	if ( FindNumbersWithSum(a,6,15,&num1,&num2) )
	{
		cout <<"num1 = "<<num1<<"  num2 = "<<num2<<endl;
	}
	else
		cout <<"not found!"<<endl;

	return 0;
}
 

分析：與上述思路基本類似，把small，big分別初始化為1,2。例如，我們求和為9的全序列，那麼small和big的變化如下圖所示。

程式碼：

#include "stdafx.h"
#include <iostream>
using namespace std;

void PrintContinuousSequence(int small, int big)
{
	cout <<"Continusou Sequence is:  ";
	for (; small <= big; small++)
		cout << small <<" ";
	cout << endl;
}

void FindContinuousSequence(int sum)
{
	if (sum < 3)
		return;
	int small = 1;
	int big = 2;
	int middle = (1+sum)>>1;
	int curSum = small + big;

	while ( small < middle )
	{
		if ( curSum == sum )
		{
			PrintContinuousSequence(small,big);
			big++;
			curSum +=big;
		}
		else if ( curSum > sum )
		{
			curSum -= small;
			small++;
		}
		else
		{
			big++;
			curSum += big;
		}
	}
}

int _tmain(int argc, _TCHAR* argv[])
{
	FindContinuousSequence(15);

	return 0;
}