Search for a Range leetcode

最新推荐文章于 2018-11-13 00:24:26 发布

原创最新推荐文章于 2018-11-13 00:24:26 发布 · 568 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种在已排序数组中查找特定目标值的高效算法，其运行复杂度为O(log n)，并提供了实现代码。

Search for a Range

Mar 3 '12

5829 / 15540

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[] A, int target) {
        int[] res = {-1,-1};
        if(A==null||A.length==0) return res;
        
        
        
        int start = 0,end = A.length-1;
        
        if(A[0]==target)  res[0]=0;
        else{
            while(start<=end){
                int mid  = start+(end-start)/2;
                if(mid-1>=0 && A[mid-1]<target &&  A[mid]==target){
                    res[0]=mid;
                    break;
                }else if(A[mid]>=target)  end = mid-1;
                else start = mid+1;
            }
        }
           
        start = 0;
        end = A.length-1;   
        if(A[A.length-1]==target) res[1]=A.length-1;
        else{
            while(start<=end){
                int mid  = start+(end-start)/2;
                if(mid+1<A.length && A[mid+1]>target &&  A[mid]==target){
                    res[1]=mid;
                    break;
                }else if(A[mid]>target)  end = mid-1;
                else start = mid+1;
            }
        }
        return res;
    }
}