LeetCode: Search for a Range 解题报告

Search for a Range
Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm‘s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

SOLUTION 1:

使用改进的二分查找法。终止条件是：left < right - 1 这样结束的时候，会有2个值供我们判断。这样做的最大的好处是，不用处理各种越界问题。

感谢黄老师写出这么优秀的算法：http://answer.ninechapter.com/solutions/search-for-a-range/

请同学们一定要记住这个二分法模板，相当好用哦。

1. 先找左边界。当mid == target,将right移动到mid，继续查找左边界。

最后如果没有找到target,退出

2. 再找右边界。当mid == target,将left移动到mid，继续查找右边界。

最后如果没有找到target,退出

 1 public class Solution {
 2     public int[] searchRange(int[] A, int target) {
 3         int[] ret = {-1, -1};
 4
 5         if (A == null || A.length == 0) {
 6             return ret;
 7         }
 8
 9         int len = A.length;
10         int left = 0;
11         int right = len - 1;
12
13         // so when loop end, there will be 2 elements in the array.
14         // search the left bound.
15         while (left < right - 1) {
16             int mid = left + (right - left) / 2;
17             if (target == A[mid]) {
18                 // 如果相等，继续往左寻找边界
19                 right = mid;
20             } else if (target > A[mid]) {
21                 // move right;
22                 left = mid;
23             } else {
24                 right = mid;
25             }
26         }
27
28         if (A[left] == target) {
29             ret[0] = left;
30         } else if (A[right] == target) {
31             ret[0] = right;
32         } else {
33             return ret;
34         }
35
36         left = 0;
37         right = len - 1;
38         // so when loop end, there will be 2 elements in the array.
39         // search the right bound.
40         while (left < right - 1) {
41             int mid = left + (right - left) / 2;
42             if (target == A[mid]) {
43                 // 如果相等，继续往右寻找右边界
44                 left = mid;
45             } else if (target > A[mid]) {
46                 // move right;
47                 left = mid;
48             } else {
49                 right = mid;
50             }
51         }
52
53         if (A[right] == target) {
54             ret[1] = right;
55         } else if (A[left] == target) {
56             ret[1] = left;
57         } else {
58             return ret;
59         }
60
61         return ret;
62     }
63 }