LeetCode-Search for a Range

題目

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].

分析

已經排好了序，用二分查詢。

C++實現1

class Solution {
public:
	vector<int> searchRange(int A[], int n, int target) {
		const int l = distance(A, lower_bound(A, A + n, target));
		const int u = distance(A, prev(upper_bound(A, A + n, target)));
		if (A[l] != target) // not found
			return vector<int> { -1, -1 };
		else
			return vector<int> { l, u };
		}
};

 

C++實現2

#include <iostream>
#include <vector>
#include <iterator>
using namespace std;


int lower_bound(vector<int>& arr, int key) {
    int half;
    int len = arr.size();
    int mid;
    int first = 0;
    while (len > 0) {
        half = len >> 1;
        mid = first + half;
        //in the right part
        if (arr[mid] < key) {
            first = mid + 1;
            //因為first=mid+1,所以這裡的len需要在減去half的基礎之上再減去1
            len = len - half - 1;
        } else {
            //in the left part
            len = half;
        }
    }
    return first;
}
int upper_bound(vector<int>& arr, int key) {
    int mid;
    int first = 0;
    int len = arr.size();
    int half;
    while (len > 0) {
        half = len >> 1;
        mid = half + first;
        if (arr[mid] > key) {//in the left part
            len = half;
        } else {//if arr[mid]<= key ,in the right part
            first = mid + 1;
            len = len - half - 1;
        }
    }
    return first;
}
	
class Solution {
public:
	vector<int> searchRange (vector<int>& arr,int target) {
		int lower = lower_bound(arr, target);
		int upper = upper_bound(arr, target);
		vector<int> res; 
		if (lower != -1)
			res.push_back(lower);
		res.push_back(upper-1);
		return res;
	}

};


int main(int argc, char** argv) {
	Solution sol;
	vector<int> arr1;
	arr1.push_back(5);
	arr1.push_back(7);
	arr1.push_back(7);
	arr1.push_back(8);
	arr1.push_back(8);
	arr1.push_back(10);
	
	int target = 8;
	vector<int> v =sol.searchRange(arr1,target);
	vector<int>::iterator it;
	for(it=v.begin(); it!=v.end();it++){	
		cout<<*it<<" "<<endl;	
	}
	system("pause"); 
	return 0;
}
 

備註

lower_bound演算法要求在已經按照非遞減順序排序的陣列中找到第一個大於等於給定值key的那個數，其基本實現原理是二分查詢。
　upper_bound函式要求在按照非遞減順序排好序的陣列中找到第一個大於給定值key的那個數，其基本實現原理是二分查詢。
兩種實現參考了stl中的實現方式，返回滿足條件的值在陣列中的下標。如果找不到滿足條件的值，將會返回陣列的大小，就像迭代器中的end一樣，對應有效下標的下一個值。