[leetcode]34.Find First and Last Position of Element in Sorted Array找区间

Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

题意：

给定一个有序数组，找出某个值的起始和终止区间。

思路：

二分查找

代码：

 class Solution {
     public int[] searchRange(int[] nums, int target) {
         // corner case
         if(nums == null || nums.length == 0) return new int[]{-1,-1};
         int left = findFirst(nums, 0, nums.length-1, target);
         if(left == -1) return new int[]{-1,-1};
         int right = findLast(nums, 0, nums.length-1, target);
         return new int[]{left, right};

     }
     // find start point
     private int findFirst(int[] nums, int left, int right, int target) {
         // avoid dead looping
         while(left + 1 < right){
             // avoid overflow
             int mid =  left + (right - left)/2;
             //   left------ |mid| ---target---right
             if(nums[mid] < target){
                 left = mid;
             }
             //   left---target---|mid| ------right
             else{
                 right = mid;
             }
         }
         if (nums[left] == target) return left;
         if (nums[right] == target) return right;
         return -1;
     }

     private int findLast(int[] nums, int left, int right, int target) {
         while(left + 1 < right){
             int mid =  left + (right - left)/2;
              //   left---target---|mid| ------right
             if(nums[mid] > target){
                 right = mid;
             }
             //   left------ |mid| ---target---right
             else{
                 left = mid;
             }
         }
         if(nums[right] == target) return right;
         if (nums[left] == target) return left;
         return -1;
     }
 }