LeetCode----- 34.Search for a Range

Given an array of integers sorted in ascending order, 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].

给定按升序排序的整数数组，找到给定目标值的开始和结束位置。算法的时间复杂度为O（log n）。如果在数组中找不到目标，则返回[-1，-1]。

代码如下:

public class SearchforARange {

	public static void main(String[] args) {
		int[] nums = new int[]{5, 7, 7, 8, 8, 10};
		int[] searchRange = searchRange(nums, 8);
		for (int i = 0; i < 2; i++) {
			System.out.print(searchRange[i]+",");
		}
	}
	
    public static  int[] searchRange(int[] nums, int target) {
    	int[] result = new int[]{-1,-1};
    	if(nums.length < 0) {
    		return result;
    	}
    	int start=0;
    	int end = nums.length-1;
    	for(;start<nums.length;start++) {
    		if(nums[start] == target) {
    			result[0]=start;
    			break;
    		}
    	}
    	for(;end>=start;end--){
    		if(nums[end] == target) {
    			result[1]=end;
    			break;
    		}
    	}
    	return result;
    }
}