searchRange.java

import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.Arrays;


/*给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值，返回 [-1, -1]。

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]
示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。*/


public class Solution {



    public static void main(String[] args){
        int nums[] = {5,7,7,8,8,10};
                int target = 8;
        System.out.print("result="+ Arrays.toString(searchRange(nums,target)));
    }
	

   private static int binarySearch(int[] nums, int target, boolean searchLeft,int startIndex) {
        int leftIndex = startIndex;
        int rightIndex = nums.length;
        int cur = 0;
        while (leftIndex < rightIndex) {
            cur = (leftIndex + rightIndex) / 2;
            if (nums[cur] > target || (nums[cur] == target && searchLeft)) {
                rightIndex = cur;
            } else {
                leftIndex = cur + 1;
            }
        }
        return leftIndex;
    }


    public static int[] searchRange(int[] nums, int target) {
        int position[]={-1,-1};
        int pre = binarySearch(nums,target,true,0);
        if (pre==nums.length || nums[pre]!=target) {
            return position;
        }
        position[0]=pre;
        position[1]=binarySearch(nums,target,false,pre)-1;        
        return position;


    }


}