给定一个数组，找到所有对号的总结给定值。 有保持2指针在正面和背面，并使他们更加接近找到一对经典的O（n）的算法。这只会导致1对。如果你希望所有对。 奖励：找到的最小距离对

您可以做到这一点为O（n）。

  INT ARR [SIZE] / *按升序排序* /

 诠释A = 0，B =大小 -  1，mindist = -1，总和;
 而（A＆LT; B）{
    总和= ARR [α] +改编并[b];
    如果（总和== TARGET）{report_pair（A，B）; mindist = B  -  A; A ++}
    否则，如果（总和＆lt;目标）A ++;
    其他b--;
 }

 / *最小距离储存在mindist * /
 

Given an array, find all pair of nos that sum up to a given value. There is the classic O(n) algorithm of keeping 2 pointers at the front and back and bringing them closer to find the pair. This only leads to 1 pair. What if you want all pairs. Bonus: Find the minimum distance pair.

Can you do this in O(n).

 int arr[SIZE]; /* sorted in ascending order */

 int a = 0, b = SIZE - 1, mindist = -1, sum;
 while (a < b) {
    sum = arr[a] + arr[b];
    if (sum == TARGET) { report_pair(a, b); mindist = b - a; a++ }
    else if (sum < TARGET) a++;
    else b--;
 }

 /* minimum distance stored in mindist */