[binary search - simple] sword finger offer II 068 Find insertion location

【 subject 】
【 Example 】
Example 1:

Input : nums = [1,3,5,6], target = 5
Output : 2

Example 2:

Input : nums = [1,3,5,6], target = 2
Output : 1

Example 3:

Input : nums = [1,3,5,6], target = 7
Output : 4

Example 4:

Input : nums = [1,3,5,6], target = 0
Output : 0

Example 5:

Input : nums = [1], target = 0
Output : 0

【 Tips 】

1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums  Arrange the array in ascending order without repeating elements 
-104 <= target <= 104

【 Code 】
Execution time ：32 ms, In all Python3 Defeated in submission 87.47% Users of
Memory consumption ：15.6 MB, In all Python3 Defeated in submission 34.53% Users of
Pass the test case ：62 / 62

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