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Lintcode 14 First Position of Target

时间:2016-01-20 20:41:52      阅读:241      评论:0      收藏:0      [点我收藏+]

Q: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.

A: binary search

 

	   public int binarySearch(int[] nums, int target){
	     if (nums == null || nums.length == 0) return -1;
	     int start = 0;
	     int end = nums.length - 1;
	     while (start + 1 < end){
	       int mid = start + (end - start) / 2;
	       if (nums[mid] >= target) end = mid;
	       else start = mid;
	     }
	     if (nums[start] == target) return start;
	     else if (nums[end] == target) return end;
	     else return -1;
	   }

  

  

 

Lintcode 14 First Position of Target

原文:http://www.cnblogs.com/nobody2somebody/p/5146293.html

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