First Position of Target

First Position of Target

给你一个排序好的数组 之后给你一个数字返回第一个出现的数组位置的下标 要求O(log n)

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Example

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

 

首先想到的肯定就是二分 因为数组有序

 

class Solution{
public:
    int binarySearch(std::vector<int> &array,int target){
        int l = 0;
        int r  = array.size()-1;
        int mid;
        while(l<=r){
        mid =l+(r-l)/2;
        if(target==array[mid]){
            if(mid==0)
                return mid;
        }
        else if (array[mid]!=array[mid-1])
        {
            return mid;
        }
        else{
            r = mid -1;
        }
        else if(target<array[mid]){
            r=mid+1；
        }
        else{
            l = mid+1;
        }
        }
    return -1;
}
};