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 }

View Code

 

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/divide2/SearchRange.java