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  • leetcode--Search for a Range

    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].

    The following code needs O(n) for the worst case.

    public class Solution {
        public int[] searchRange(int[] A, int target) {
            int[] indices = new int[]{-1,-1};
            int length = A.length;
            int index = Arrays.binarySearch(A, target);
            if(index >= 0){
            	int left = index, right = index;
            	while(left > -1 && A[left] == target){
                    --left;
            	}
            	while(right < length && A[right] == target){
            	    ++right;
            	}
            	indices[0] = left + 1;
            	indices[1] = right - 1;
            }
            return indices;    
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/averillzheng/p/3783165.html
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