[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;
    }
}