First Position of Target

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.

分析：
很明显的用binary search, 但是因为要找第一个，所以，我们需要判断一下。

 class Solution {
     /**
      * @param nums: The integer array.
      * @param target: Target to find.
      * @return: The first position of target. Position starts from 0.
      */
     public int binarySearch(int[] nums, int target) {
         if (nums == null || nums.length == ) return -;

         int start = ;
         int end = nums.length - ;

         while (start <= end) {
             int mid = start + (end - start) / ;
             if (nums[mid] == target) {
                 if (mid !=  && nums[mid - ] == target) {
                     end = mid - ;
                 } else {
                     return mid;
                 }
             } else if (nums[mid] < target) {
                 start = mid + ;
             } else {
                 end = mid - ;
             }
         }
         return -;
     }
 }

参考请注明出处： cnblogs.com/beiyeqingteng/