和为s的两个数字VS 和为S的连续正数序列

输入一个递增排序的数组和一个数字是S，在数组中查找两个数，使得它们的和正好是S。

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

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

    while(ahead  > behind)
    {
        long curSum = data[ahead]+data[behind];

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

    return found;
}

主要运用前后2个指针（代码中没写成指针形式）ahead,behind减少了遍历次数，增加了效率。

 

输入一个正数S，打印所有和为S的连续正数序列。思路是相同的

void PrintContinuousSequence(int small,int big)
{
    for(int i =small; i <= big; ++i)
    {
        cout<<i <<'\t';
    }
    cout<<endl;
}


void FindContinuousSequence(int sum)
{
    if( sum < 3)
        return;

    int small = 1;
    int big = 2;
    int middle = (1 + sum)/2;
    int curSum = small + big;

    while( small < middle)
    {
        if(curSum == sum)
            PrintContinuousSequence(small,big);

        while( curSum > sum && small <middle)
        {
            curSum -=small;
            small++;

            if(curSum == sum)
                PrintContinuousSequence(small,big);
        }
        big ++;
        curSum +=big;
    }
}