CareerCup Google Interview 找到两个数的差为一个特定数

Given an integer 'k' and an sorted array A (can consist of both +ve/-ve nos), output 2 integers from A such that a-b=k.
PS:
nlogn solution would be to check for the occurence of k-a[i] (using binary search) when you encounter a[i]. methods like hash consume space.

Is an O(n) solution with O(1) extraspace possible?

http://www.careercup.com/question?id=9449188

利用2sum的思想，设定两个指针。由于2sum是在数组两端设定指针，而对于difference，在两端设定指针当a[j] - a[i] > key时， j--或者i++都会令a[j]-a[i]的差变小，不具有唯一性。

所以最开始设定指针都在同一端，这样a[j]-a[i]>key时，只有i++一个操作能令差变小。同样a[j]-a[i]<key时，也只有j++能令差变大。当指针相遇，或者j==a.size()时说明没有解。同时，由于数组中可能有负数，且key可能为负数，所以用abs(key)来进行比较。

#include <iostream>
#include <vector>
using namespace std;

void solve(vector<int> &a, int key)
{
    if (a.size() <= 1)
    {
        cout << "No answer!" << endl;
        return;
    }

    int i = 0;
    int j = 1;

    key = abs(key);

    while(i < j && j < a.size())
    {
        if (a[j] - a[i] == key)
        {
            cout << a[j] << ' ' << a[i] << endl;
            return;
        }
        else if (a[j] - a[i] > key)
        {
            i++;
        }
        else
        {
            j++;
        }
    }

    cout << "No answer!" << endl;
}

int main()
{
    vector<int> a;
    for(int i = 0; i < 10; i++)
        a.push_back(i);

    solve(a, 1);
    solve(a, 2);
    solve(a, 9);
    solve(a, 10);
    solve(a, -1);
}