leetcode116:search-for-a-range

题目描述

给出一个有序数组，请在数组中找出目标值的起始位置和结束位置

你的算法的时间复杂度应该在O(log n)之内

如果数组中不存在目标，返回[-1, -1].

例如：

给出的数组是[5, 7, 7, 8, 8, 10]，目标值是8,

返回[3, 4].

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].

示例1

输入

复制

[5, 7, 7, 8, 8, 10],8

输出

复制

[3,4]


class Solution {
public:
    /**
     * 
     * @param A int整型一维数组 
     * @param n int A数组长度
     * @param target int整型 
     * @return int整型vector
     */
    vector<int> searchRange(int* A, int n, int target) {
        // write code here
        vector< int> res(2,-1);
        if (A==nullptr || n<=0)
            return res;
        int low=lower_bound(A, A+n, target)-A;
        if (low==n || A[low]!=target)
            return res;
        else res[0]=low;
        int high=upper_bound(A, A+n,target)-A-1;
        res[1]=high;
        return res;
        
    }
};

class Solution {
public:
    /**
     * 
     * @param A int整型一维数组 
     * @param n int A数组长度
     * @param target int整型 
     * @return int整型vector
     */
    vector<int> searchRange(int* A, int n, int target) {
        // write code here
        vector<int> res(2,-1);
        if (A==nullptr || n<=0)
            return res;
        int low=0,high=n-1;
        while (low<=high)
        {
            int middle =(high+low)>>1;
            if (A[middle]<target)
                low=middle+1;
            else 
                high=middle-1;
            
        }
        int low2=0,high2=n-1;
        while (low2<=high2)
        {
            int middle2=(high2+low2)>>1;
            if (A[middle2]<=target)
                low2=middle2+1;
            else 
                high2=middle2-1;
            
        }
        if (low<=high2){
            res[0]=low;
            res[1]=high2;
            
        }
        return res;
    }
    
};