[Swift]LeetCode964. 表示数字的最少运算符 | Least Operators to Express Number
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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 expressions 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
给定一个正整数 x
,我们将会写出一个形如 x (op1) x (op2) x (op3) x ...
的表达式,其中每个运算符 op1
,op2
,… 可以是加、减、乘、除(+
,-
,*
,或是 /
)之一。例如,对于 x = 3
,我们可以写出表达式 3 * 3 / 3 + 3 - 3
,该式的值为 3 。
在写这样的表达式时,我们需要遵守下面的惯例:
- 除运算符(
/
)返回有理数。 - 任何地方都没有括号。
- 我们使用通常的操作顺序:乘法和除法发生在加法和减法之前。
- 不允许使用一元否定运算符(
-
)。例如,“x - x
” 是一个有效的表达式,因为它只使用减法,但是 “-x + x
” 不是,因为它使用了否定运算符。
我们希望编写一个能使表达式等于给定的目标值 target
且运算符最少的表达式。返回所用运算符的最少数量。
示例 1:
输入:x = 3, target = 19 输出:5 解释:3 * 3 + 3 * 3 + 3 / 3 。表达式包含 5 个运算符。
示例 2:
输入:x = 5, target = 501 输出:8 解释:5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5 。表达式包含 8 个运算符。
示例 3:
输入:x = 100, target = 100000000 输出:3 解释:100 * 100 * 100 * 100 。表达式包含 3 个运算符。
提示:
2 <= x <= 100
1 <= target <= 2 * 10^8
100ms
1 class Solution { 2 var x:Int = 0 3 var best:Int = 0 4 func leastOpsExpressTarget(_ x: Int, _ target: Int) -> Int { 5 var target = target 6 self.x = x 7 var list:[Int] = [Int]() 8 while(target != 0) 9 { 10 list.append(target % x) 11 target /= x 12 } 13 self.best = Int.max 14 dfs(list, 0, 0, 0) 15 return best - 1 16 } 17 18 func dfs(_ list:[Int],_ k:Int,_ add:Int,_ count:Int) 19 { 20 var add = add 21 if count >= best 22 { 23 return 24 } 25 if add == 0 && k >= list.count 26 { 27 best = min(best, count) 28 return 29 } 30 31 if k < list.count 32 { 33 add += list[k] 34 } 35 var cost:Int = k == 0 ? 2 : k 36 var cur:Int = add % x 37 add /= x 38 dfs(list, k + 1, add, count + cost * cur) 39 if cur != 0 && k <= list.count + 2 40 { 41 dfs(list, k + 1, 1, count + cost * (x - cur)) 42 } 43 } 44 }