Java_LeetCode34~Find First and Last Position of Element in Sorted Array

最新推荐文章于 2022-03-31 10:19:50 发布

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最新推荐文章于 2022-03-31 10:19:50 发布

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博客围绕在升序排序的整数数组中查找给定目标值的起始和结束位置展开。要求算法时间复杂度为O(log n)，若未找到目标值则返回 [-1, -1]。采用二分查找左右范围，当中间值等于目标值时继续在相应区域查找。

Find First and Last Position of Element in Sorted Array

Description

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]

Solution

To search left&right range with binaryAG ;When mid equals targert,keep searching the left area in case of leaving THE accurate left range;So is the right;

class Solution {
    public int[] searchRange(int[] nums, int target) {
        return new int[]{
            binarySearch(nums, target, true), 
            binarySearch(nums, target, false)
        };
    }
    
    private int binarySearch(int[] nums, int target, boolean isLeft) {
        int low = 0;
        int high = nums.length - 1;
        int index = -1;
        while (low <= high) {
            int mid = (low + high) / 2;
            
            if (nums[mid] < target)
                low = mid + 1;                
            else if (nums[mid] > target)
                high = mid - 1;
            else {
                index = mid;
                if (isLeft)
                    high = mid - 1;
                else
                    low = mid + 1;
            }
        }
        return index;
    }
}