Lintcode: Binary Search

Binary search is a famous question in algorithm.



For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.



If the target number does not exist in the array, return -1.



Example

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.



Challenge

If the count of numbers is bigger than MAXINT, can your code work properly?

跟Leetcode里search for a range挺像的，就是找到一个target之后，还要继续找它的左边沿。最后l指针超过r指针之后， l 指针会停在左边沿上

 1 class Solution {

 2     /**

 3      * @param nums: The integer array.

 4      * @param target: Target to find.

 5      * @return: The first position of target. Position starts from 0.

 6      */

 7     public int binarySearch(int[] nums, int target) {

 8         int l = 0, r = nums.length - 1;

 9         int m = 0;

10         while (l <= r) {

11             m = (l + r) / 2;

12             if (nums[m] == target) break;

13             else if (nums[m] > target) r = m - 1;

14             else l = m + 1;

15         }

16         if (nums[m] != target) return -1;

17         l = 0;

18         r = m;

19         while (l <= r) {

20             m = (l + r) / 2;

21             if (nums[m] == target) {

22                 r = m - 1;

23             }

24             else l = m + 1;

25         }

26         return l;

27     }

28 }