Codility算法测验(一)

    几天前有人给了Codility上的三个算法，本人向来脑子不灵光，加上这么多年没弄过算法了，感觉有些吃力。这些算法的solution自知很不成熟，希望抛砖引玉，多提意见。另外为避免翻译不当，令大家会错了意，故将原文贴出。

1. You want to park your bicycle in a bicycle parking area where bike racks are aligned in a row. There are already N bikes parked there(each bike is attached to exactly one rack, but a rack can have multiple bikes attached to it). We call racks that already have bikes attached used.
    You want to park your bike in a rack in the parking area according to the following criteria:
           the chosen rack must lie between te first and the last used racks(inclusive);
           the distance between the chose rack and any other used rack is as big as possible.
   A description of the bikes already parked in the racks is given in a non-empty zero-indexed array A: element A[K] denotes the postion of the rack to which bike number K is attaced (for 0 <= K < N). The central position in the parking area is position 0. A positive value means that the rack is located A[K] meters to the right of the cetral position 0; a negative value means that it's located |A[K]| meters to the left.
   That, given a non-empty zero-indexed array A of N integers, returns the largest possible distance in meters between the chosen rack and any other used rack.
   Assume that:
           N is an integer within the range [2..100,000];
           each element of array A is an integer within the range [-1,000,000,000..1,000,000,000].
   Complexity:
           expected worst-case time complexity isO(N*log(N));
           expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

   思考： 我觉得题目主要是篇阅读理解，实际上就是要寻求两点的最大距离，然后将车子停在这两点最中间的位置

    

2. public int solution(int[] A) {
       Arrays.sort(A);
       int ans = Integer.MIN_VALUE;
       if (A.length == 2) return (A[1] - A[0]) / 2;
       for (int i=0; i < A.length - 1; i++) {
           //如果两点之间有空位置
           if (A[i+1] - A[i] > 1) {
               ans = Math.max(ans, A[i+1] - A[i]);
           }
       }
           
       return ans / 2;
   }