SearchForARange.java ( File view )

  • By issaczr 2014-10-24
  • View(s):52
  • Download(s):0
  • Point(s): 1
			// Author:   Li Long, 1988lilong@163.com
// Date:     Apr 17, 2014
// Source:   http://oj.leetcode.com/problems/search-for-a-range/ 
// Analysis: http://blog.csdn.net/lilong_dream/article/details/22893675

// 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]. 

public class SearchForARange {

	public int[] searchRange(int[] A, int target) {

		int left = 0;
		int right = A.length - 1;

		int[] result = {
 -1, -1 
};

		while (left <= right) {

			int mid = (left + right) / 2;

			if (A[mid] > target) {

				right = mid - 1;
			
} else if (A[mid] < target) {

				left = mid + 1;
			
} else {

				result[0] = mid;
				result[1] = mid;

				int i = mid - 1;
				while (i >= 0 && A[i] == target) {

					result[0] = i;
					--i;
				
}

				i = mid + 1;
				while (i < A.length && A[i] == target) {

					result[1] = i;
					++i;
				
}

				break;
			
}
		
}

		return result;
	
}
	
	public static void main(String[] args) {

		int[] A = {
 1, 3, 4, 5, 5 
};

		SearchForARange slt = new SearchForARange();
		int[] result = slt.searchRange(A, 5);
		System.out.println(result[0] + " " + result[1]);
	
}

}			
...
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