Search for a Range !!!

最新推荐文章于 2022-03-08 23:18:06 发布

benbenab

最新推荐文章于 2022-03-08 23:18:06 发布

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分类专栏： number process

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本博客介绍了一种O(log n)复杂度的方法来查找有序数组中给定目标值的起始和结束位置。如果目标不存在，则返回[-1, -1]。

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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].

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> result(2,-1);
        
        
        
    
        int low = FindLowerBound(A,0, n-1 ,target,n);
        low = low + 1; //move to next element

    	if(A[low] == target)
           result[0] = low;
        else
            return result;
		int upper = FindUpperBound(A,0, n-1 ,target,n);
		 upper = upper < 0? (n-1):(upper - 1);
		 result[1] = upper;


        return result;
    }
    
    int FindUpperBound(int A[], int start, int end, int target,int n){
        if (start > end || A[end] <= target)
            return -1;
        if (start == end)
            return start;

        int mid = (start+end)/2;
        if (A[mid] <= target){
			if (mid > 0 && A[mid-1] == target)
                return mid;
            else
                FindUpperBound(A,mid+1, end ,target,n);
        }
        else
            FindUpperBound(A,start, mid ,target,n);
    }
    
    int FindLowerBound(int A[], int start, int end, int target, int n){
        if (start > end|| A[start] >= target)
            return -1;
        if (start == end)
            return start;
            
        int mid = (start+end+1)/2;
        if (A[mid] >= target ){
            if (mid+1<n && A[mid+1] == target )
                return mid;
            else
                FindLowerBound(A, start, mid-1, target,n);
        }
            
        else{
            FindLowerBound(A, mid, end, target,n);
        }
    }
};