[剑指offer]和为s的两个数 ； 和为s的连续正整数序列

/*    41：和为s的两个数 ； 和为s的连续正整数序列    {1，2，4，7，11，15}  求和为15的两个数(头尾指针)*/bool FindTwoNumOfSum(int ar[], int len, int sum, int *num1, int *num2){       bool flag = false;    if (ar == NULL || len <= 0 || num1 == NULL || num2 == NULL)        return flag;    int head = 0;    int tail = len - 1;    while (head < tail)    {        int curSum = ar[head] + ar[tail];        if (curSum == sum)        {            *num1 = ar[head];            *num2 = ar[tail];            flag = true;            break;        }        else if (curSum>sum)            tail--;        else            head++;    }    return flag;}/*    打印和为n的序列    1+2+3+4+5=4+5+6=7+8=15    打印 1~5，4~6，7~8*/void printSeq(int small, int big){    for (int i = small; i <= big; ++i)        cout << i << " ";    cout << endl;}void FindSeqOfSum(int sum){    if (sum < 3)        return;    int small = 1;    int big = 2;    int times = (sum + 1) / 2; //small big两个数  只用增加到sum的一半    int curSum = small + big;    while (small < times)    {        if (curSum == sum)            printSeq(small,big);        while (curSum > sum && small < times)        {            curSum -= small;            small++;            if (curSum == sum)                printSeq(small, big);        }        big++;        curSum += big;    }}////void test()//{//  FindSeqOfSum(15);//}