14. First Position of Target 【easy】

For a given sorted array (ascending order) and a targetnumber, 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 2^32, can your code work properly?

解法一：

 1 class Solution {
 2 public:
 3     /**
 4      * @param nums: The integer array.
 5      * @param target: Target number to find.
 6      * @return: The first position of target. Position starts from 0. 
 7      */
 8     int binarySearch(vector<int> &array, int target) {
 9         if (array.size() == 0) {
10             return -1;
11         }
12         
13         int start = 0;
14         int end = array.size() - 1;
15         
16         while (start + 1 < end) {
17             int mid = start + (end - start) / 2;
18             
19             if (array[mid] == target) {
20                 end = mid;
21             }
22             else if (array[mid] < target) {
23                 start = mid;
24             }
25             else if (array[mid] > target) {
26                 end = mid;
27             }
28         }
29         
30         if (array[start] == target) {
31             return start;
32         }
33         
34         if (array[end] = target) {
35             return end;
36         }
37         
38         return -1;
39     }
40 };

解法二：

 1 class Solution {
 2 public:
 3 
 4     int find(vector<int> &array, int start, int end, int target) {
 5         if (start > end) {
 6             return -1;
 7         }
 8         
 9         int mid = start + (end - start) / 2;
10         
11         if (array[mid] == target) {
12 
13             if (array[mid - 1] != target) {
14                 return mid;
15             }
16 
17             return find(array, start, mid - 1, target);
18         }
19         else if (array[mid] > target) {
20             return find(array, start, mid - 1, target);
21         }
22         else if (array[mid] < target) {
23             return find(array, mid + 1, end, target);
24         }
25         
26     }
27 
28     /**
29      * @param nums: The integer array.
30      * @param target: Target number to find.
31      * @return: The first position of target. Position starts from 0. 
32      */
33     int binarySearch(vector<int> &array, int target) {
34         // write your code here
35         
36         int start = 0;
37         int end = array.size();
38         
39         return find(array, start, end, target);
40         
41     }
42 };