Binary Search（Easy，二分搜索）

题目描述：

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.

 

样例：

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

挑战：

If the count of numbers is bigger than 2^32, can your code work properly?

code：

int binarySearch(vector<int> &array, int target) {
        // write your code here
        long long first = 0; //防止数据过大，用long long类型
        long long nums = array.size();
        long long last = nums-1;
        long long mid;
        while(first <= last){
            mid = first + ((last - first + 1) >> 1);//防止数据过大，用移位
            if(array[mid] == target){
                if(mid == first || array[mid] > array[mid - 1])
                    return mid;//相等时，当mid等于first或者mid指向的位置大于前一个位置才能说明是首个
                else if(array[mid] == array[mid - 1])//否则继续二分，找到第一个
                    last = mid -1;
            }else if(array[mid] > target){
                last = mid - 1;
            }else{
                first = mid + 1;
            }
        }
        return -1;
    }