leetcode -- 34. Find First and Last Position of Element in Sorted Array

最新推荐文章于 2020-08-08 23:55:38 发布

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本文介绍了一种中等难度的算法问题，即在升序整数数组中查找给定目标值的开始和结束位置，同时确保算法的时间复杂度为O(logn)。提供了两种解决方案，包括遍历查找和二分查找优化的方法。

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题目描述

题目难度：Medium
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]

AC代码1

这个题感觉没什么亮点，不做也罢

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = new int[]{-1,-1};
        if(nums == null || nums.length == 0) return res;
        for(int i = 0;i < nums.length;i++) 
            if(nums[i] == target) {
                res[0] = i;
                break;
            }
        for(int i = nums.length - 1;i >= 0;i--) 
            if(nums[i] == target) {
                res[1] = i;
                break;
        }
        return res;
    }
}

AC代码2

class Solution {
    public int[] searchRange(int[] nums, int target) {
        
        int lo=0;
        int hi = nums.length;
        
        int[] targetRange = {-1, -1};
        
        boolean found = false;
        int mid = -1;
        while(lo<hi){
            mid = (lo+hi)/2;
            if(target == nums[mid]){
                found = true;
                break;
            }
            
            if(nums[mid]>target){
                hi = mid;
            } else {
                lo = mid+1;
            }
        }
        
        if(found){
            
            int left = mid;
            while(mid>=0 && mid < nums.length  && nums[mid]==target){
                mid--;
            }
            
            targetRange[0]=mid+1;
            mid=left;
            while(mid>=0 && mid < nums.length && nums[mid]==target){
                mid++;
            }
            targetRange[1]=mid-1;
        }
        
        return targetRange;
    }

}