Count of Smaller Number After Itself

 1 public class Solution {
 2     public List<Integer> countSmaller(int[] nums) {
 3         List<Integer> result = new ArrayList<>();
 4         List<Integer> record = new ArrayList<>();
 5         
 6         if (nums.length == 0) {
 7             return result;
 8         }
 9         result.add(0);
10         record.add(nums[nums.length - 1]);
11         for (int i = nums.length - 2; i >= 0; i--) {
12             int start = 0, end = result.size() - 1, mid = 0;
13             while (start <= end) {
14                 mid = (end - start)/2 + start;
15                 if (record.get(mid) >= nums[i]) {
16                     end = mid - 1;
17                 } else {
18                     start = mid + 1;
19                 }
20             }
21             result.add(0, start);
22             record.add(start, nums[i]);
23         }
24         return result;
25     }
26 }

1. This binary search is find the insertion position. Since it starts from end of array. So it must be greater and equal to mid number.

The merge sort method is pretty tricky

posted on 2016-07-05 14:51 keepshuatishuati 阅读(111) 评论(0) 编辑收藏举报