Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Add solution and test-cases for problem 1404 #889

Merged
merged 1 commit into from
Jun 6, 2024
Merged
Show file tree
Hide file tree
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
Original file line number Diff line number Diff line change
@@ -0,0 +1,47 @@
# [1404.Number of Steps to Reduce a Number in Binary Representation to One][title]

## Description
Given the binary representation of an integer as a string `s`, return the number of steps to reduce it to `1` under the following rules:

- If the current number is even, you have to divide it by `2`.

- If the current number is odd, you have to add `1` to it.

It is guaranteed that you can always reach one for all test cases.

**Example 1:**

```
Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14.
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.
Step 5) 4 is even, divide by 2 and obtain 2.
Step 6) 2 is even, divide by 2 and obtain 1.
```

**Example 2:**

```
Input: s = "10"
Output: 1
Explanation: "10" corressponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.
```

**Example 3:**

```
Input: s = "1"
Output: 0
```

## 结语

如果你同我一样热爱数据结构、算法、LeetCode,可以关注我 GitHub 上的 LeetCode 题解:[awesome-golang-algorithm][me]

[title]: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one
[me]: https://github.com/kylesliu/awesome-golang-algorithm
Original file line number Diff line number Diff line change
@@ -1,5 +1,26 @@
package Solution

func Solution(x bool) bool {
return x
func Solution(s string) int {
bs := []byte(s)
steps := 0
idx := len(s) - 1
for idx >= 0 {
if string(bs[:idx+1]) == "1" {
break
}
if bs[idx] == '0' {
idx--
steps++
continue
}
steps += 1
for ; idx >= 0 && bs[idx] != '0'; idx-- {
steps++
}
if idx >= 0 {
bs[idx] = '1'
}
}

return steps
}
Original file line number Diff line number Diff line change
Expand Up @@ -10,12 +10,12 @@ func TestSolution(t *testing.T) {
// 测试用例
cases := []struct {
name string
inputs bool
expect bool
inputs string
expect int
}{
{"TestCase", true, true},
{"TestCase", true, true},
{"TestCase", false, false},
{"TestCase1", "1101", 6},
{"TestCase2", "10", 1},
{"TestCase3", "1", 0},
}

// 开始测试
Expand All @@ -30,10 +30,10 @@ func TestSolution(t *testing.T) {
}
}

// 压力测试
// 压力测试
func BenchmarkSolution(b *testing.B) {
}

// 使用案列
// 使用案列
func ExampleSolution() {
}
Loading