FB面经prepare: Count the number of Vector

给一个超级大的排好序的vector  [abbcccdddeeee]比如，要求返回[{,a}, {,b}, {,c}, {,d}, {,e}......]复杂度要优于O(N)

分析：

如果是binary search找每个char的上下界，worst case要找n次，时间复杂度O(nlogn)

所以考虑每次比较start point和start point + 2^n位置上的数，假如一样就continue，不一样就在区间里面binary search找上界，这样worst case O（N）

 package fb;

 import java.util.*;

 public class ReorganizeVector {

     public List<List<Character>> reorganize(String str) {
         List<List<Character>> res = new ArrayList<List<Character>>();
         if (str==null || str.length()==0) return res;
         int starter = 0;
         int n = 0;
         while (starter < str.length()) {
             char cur = str.charAt(starter);
             int index = starter + (int)Math.pow(2, n);
             while (index < str.length() && str.charAt(index) == cur) {
                 n++;
                 index = starter + (int)Math.pow(2, n);
             }
             if (index >= str.length())
                 index = str.length() - 1;
             int rightEdge = findRight(str, starter, index, cur);
             List<Character> newItem = new ArrayList<Character>();
             newItem.add((char)('0'+ rightEdge-starter+1));
             newItem.add(cur);
             res.add(newItem);

             starter = rightEdge + 1;
             n = 0;
         }
         return res;
     }

     public int findRight(String str, int starter, int end, char target) {
         int l = starter;
         int r = end;
         while (l <= r) {
             int m = (l + r)/2;
             if (str.charAt(m) == target)
                 l = m + 1;
             else r = m - 1;
         }
         return r;
     }

     /**
      * @param args
      */
     public static void main(String[] args) {
         // TODO Auto-generated method stub
         ReorganizeVector sol = new ReorganizeVector();
         List<List<Character>> res = sol.reorganize("abbcccddddeeeee");
         for (List<Character> line : res) {
             for (Character c : line) System.out.print(c);
             System.out.println("");
         }
     }

 }