给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。

请必须使用时间复杂度为 O(log n) 的算法。

示例 1:

输入: nums = [1,3,5,6], target = 5 输出: 2 示例 2:

输入: nums = [1,3,5,6], target = 2 输出: 1 示例 3:

输入: nums = [1,3,5,6], target = 7 输出: 4 示例 4:

输入: nums = [1,3,5,6], target = 0 输出: 0 示例 5:

输入: nums = [1], target = 0 输出: 0

class Solution {
    public int searchInsert(int[] nums, int target) {
        int left = 0,  right = nums.length - 1;
        int mid = 0;
        int num = 0;
        while(left <= right){
            mid = left + (right - left) / 2;
            num = nums[mid];
            if(num == target) return mid;
            else if(num < target) left = mid + 1;
            else right = mid - 1;
        }
        if(target < num) return mid;
        else return mid + 1;
    }
}

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        int right = 0, left = nums.size() - 1;
        int mid = 0;
        int num;
        while(right <= left){
            mid = left + (right - left) / 2;
            num = nums[mid];
            if(num == target) return mid;
            else if(num < target) right = mid + 1;
            else left = mid - 1;
        }
        if(num > target) return mid;
        else return mid + 1;
    }
};

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums) - 1
        while(left <= right):
            mid = left + (right - left) // 2
            num = nums[mid]
            if num == target:
                return mid
            elif num < target:
                left = mid + 1
            else :
                right = mid - 1
        if num > target:
             return mid
        else :
             return mid + 1;