Case Analysis of python search insertion location

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[title]

Given a sorted array and a target value, find the target value in the array and return its index. If the target value does not exist in the array, returns the position where it will be inserted sequentially.

You can assume that there are no repeating elements in the array.

Example 1:

Input: [1, 3, 5, 6], 5

Output: 2

Example 2:

Input: [1, 3, 5, 6], 2

Output: 1

Example 3:

Input: [1, 3, 5, 6], 7

Output: 4

Example 4:

Input: [1, 3, 5, 6], 0

Output: 0

[ideas]

1. Brute force solution: traversing the array to find elements.

2. Binary search: the question type is to find the first element greater than or equal to target and return its location.

Binary search generally uses common code:

I, j = 0, len (nums)-1

While i target:

J = mid-1

Else:

I = mid + 1

Return i

At the end of the # loop, the result is the operator in the formula nums [j] target, and the returned subscript, which are also parts of the general code that must be modified according to the title.

# when nums [mid] > target, the final result nums [j] = target, the final result nums [j]

< target int: l, r = 0, len(nums) - 1 # 第一个 >

Elements of = target

# nums [r] < target