Binary search

这题是基础题，但是细节挺多的，先上题目

Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.


Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1
 

Note:

You may assume that all elements in nums are unique.
n will be in the range [1, 10000].
The value of each element in nums will be in the range [-9999, 9999].

纯二分查找，主要要注意的是，在获得中间索引的那一步，两个int值相加可能会溢出，所以需要用减法做下处理。

还有就是左边和右边的索引该如何变，加上终止条件。代码如下：

class Solution {
public:
    static int Main(vector<int>& nums, int target) {
        int l = 0;
        int r = nums.size() - 1;

        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[mid] > target) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return -1;
    }
};