【LeetCode OJ 34】Search for a Range

题目链接：https://leetcode.com/problems/search-for-a-range/

题目：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]

.

解题思路：题意为在一个有序数组中寻找指定值的起始位置和结束位置，因为题目要求时间复杂度为O(log n)，故而想到使用二分查找法。

示例代码：

public class Solution 
{
	 public int[] searchRange(int[] nums, int target) 
	 {
	        int[] result=new int[]{-1,-1};
	        int start=0;
	        int end=nums.length-1;
	        while(start<=end)
	        {
	        	int middle=(start+end)>>1;
	        	int middleValue=nums[middle];
	        	if(middleValue==target)
	        	{
	        		//找到目标值后，先搜索其左边有没有与目标值相等的数，取其最左边的索引值放入result中，同理再搜索其最右边
	        		int i=middle;
	        		while((i-1)>=0&&nums[i-1]==target)
	        		{
	        			i--;
	        		}
	        		result[0]=i;
	        		int j=middle;
	        		while((j+1)<nums.length&&nums[j+1]==target)
	        		{
	        			j++;
	        		}
	        		result[1]=j;
	        		return result;
	        	}
	        	else if(target<middleValue)  
	            {  
	                //小于中值时在中值前面找  
	                end=middle-1;  
	            }  
	            else  
	            {  
	                //大于中值在中值后面找  
	                start=middle+1; 
	            }
	        
	        }
	        return result;
	 }
}