Search for a Range - Leetcode

最新推荐文章于 2018-11-25 21:42:21 发布

AngelQian2

最新推荐文章于 2018-11-25 21:42:21 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/Q_wing/article/details/43882883

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本博客介绍了一个在有序数组中查找指定目标值起始和结束位置的算法，利用二分查找优化效率至O(log n)。通过示例说明了如何实现并应用这一方法。

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public class Solution {
    public int[] searchRange(int[] A, int target) {
        int start=0, end=A.length-1, mid, pivot=0;
        while(start<=end){
            mid = (start+end)/2;
            if(A[mid] == target){
                pivot = mid;
                break;
            }else if(target > A[mid])
                start=mid+1;
            else
                end=mid-1;
        }
       <span style="background-color: rgb(255, 204, 204);"> if(A[pivot]!=target)
        return new int[]{-1,-1};</span>
        
        int start1=0, end1=pivot, mid1;
        while(start1 <= end1){
            mid1 = (start1+end1)/2;
            if(A[mid1] == target)
                end1=mid1-1;
            else
                start1=mid1+1;
        }
        int new_low = <span style="background-color: rgb(255, 255, 51);">start1</span>;
        
        int start2=pivot+1, end2 = A.length-1, mid2;
        while(start2 <= end2){
            mid2 = (start2+end2)/2;
            if(A[mid2] == target)
                start2=mid2+1;
            else
                end2=mid2-1;
        }
        int new_high = <span style="background-color: rgb(255, 255, 51);">end2</span>;
        
        return new int[]{new_low,new_high};
    }
}

分析：数组是有序的，找到一个值两边的index。O(log n) 暗示二叉搜索，这里考点就是如何设置Middle

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].