search-for-a-range

【题目描述】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].

【解题思路】这道题要求查找操作的时间复杂度为对数级，那毫无疑问就是二分了，但是不同于一般的二分，这道题要求返回的是元素索引组成的向量，而不是一个索引

【考查内容】查找

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int>result;
        if(A==NULL||n==0)
            return result;
        int left = helper1(A,n,target);
        int right = helper2(A,n,target);
        result.push_back(left);
        result.push_back(right);
        return result;
    }
    int helper1(int A[],int n,int target){
        int start = 0;
        int end = n-1;
        while(start<=end){
            int mid = start + (end - start)/2;
            if(A[mid]<target)
                start = mid+1;
            else if(A[mid]>target)
                end = mid-1;
            else if(A[mid]==target&&(A[mid-1]!=target||mid == 0))
                return mid;
            else
                end = mid -1;
        }
        return -1;
    }
    int helper2(int A[],int n,int target){
        int start =0;
        int end = n-1;
        while(start<=end){
            int mid =start +(end-start)/2;
            if(A[mid]<target)
                start = mid+1;
            else if(A[mid]>target)
                end = mid-1;
            else if(A[mid]==target&&(A[mid+1]!=target||mid == n-1))
                return mid;
            else
                start = mid +1;
        }
        return -1;
    }
};