Leetcode no. 34

最新推荐文章于 2021-08-18 11:01:56 发布

SmithZz

最新推荐文章于 2021-08-18 11:01:56 发布

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本文介绍了一种算法，用于在一个已排序的整数数组中找到给定目标值的起始和结束位置。该算法的时间复杂度为O(log n)，并提供了一个具体的Java实现案例。

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34. Search for a Range

Given a sorted array of integers, 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res= {-1, -1};
        if (nums.length==0 || nums[0]>target || nums[nums.length-1]<target) return res;
        int left=0, right= nums.length-1;
        while (left<right){
            int center= (left+right)/2;
            if (nums[center]<target) left= center+1;
            else right= center;
        }
        if (nums[left]!=target) return res;
        else res[0]=left;
        right= nums.length-1;
        while (left<right){
            int center= (left+right)/2+1;
            if (nums[center]>target) right= center-1;
            else left= center;
        }
        res[1]= left;
        return res;
    }
}