964. Least Operators to Express Number

964. Least Operators to Express Number

难度: Hard

刷题内容

原题连接

内容描述

Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where each operator op1, op2, etc. is either addition, subtraction, multiplication, or division (+, -, *, or /).  For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3.

When writing such an expression, we adhere to the following conventions:

The division operator (/) returns rational numbers.
There are no parentheses placed anywhere.
We use the usual order of operations: multiplication and division happens before addition and subtraction.
It's not allowed to use the unary negation operator (-).  For example, "x - x" is a valid expression as it only uses subtraction, but "-x + x" is not because it uses negation.
We would like to write an expression with the least number of operators such that the expression equals the given target.  Return the least number of operators used.



Example 1:

Input: x = 3, target = 19
Output: 5
Explanation: 3 * 3 + 3 * 3 + 3 / 3.  The expression contains 5 operations.
Example 2:

Input: x = 5, target = 501
Output: 8
Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5.  The expression contains 8 operations.
Example 3:

Input: x = 100, target = 100000000
Output: 3
Explanation: 100 * 100 * 100 * 100.  The expression contains 3 operations.


Note:

2 <= x <= 100
1 <= target <= 2 * 10^8

解题方案

思路 1
- 时间复杂度: O(lg(Target))- 空间复杂度: O(1)

参考donggua_fu

能AC, 但是beats 0%

```python
class Solution:
def leastOpsExpressTarget(self, x, target):
"""
:type x: int
:type target: int
:rtype: int
"""
# At this time, you can get target either by add target times x/x or
# subtract (x - target) times x/x to x
# For example, x = 3, target = 2. Then, 3/3 + 3/3 or 3 - 3/3 is possible result
if x > target:
return min(target * 2 - 1, (x - target) * 2)
if x == target: # just push x at the end
return 0
sums, times = x, 0
while sums < target: # this is gready, put as much as possible 'x'
times += 1
sums *= x
if sums == target: # one more 'x' you put, one more operator
return times
# when you have remainder, you have two choices, one is add, the other is subtract
# for example, x = 3, target = 5. Then, 5 = 3 + 2 or 5 = 9 - 4
l, r = sys.maxsize, sys.maxsize

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