Question

1342. Number of Steps to Reduce a Number to Zero

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Description

Given an integer num, return _the number of steps to reduce it to zero_.

In one step, if the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

 

Example 1:

Input: num = 14
Output: 6
Explanation: 
Step 1) 14 is even; divide by 2 and obtain 7. 
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3. 
Step 4) 3 is odd; subtract 1 and obtain 2. 
Step 5) 2 is even; divide by 2 and obtain 1. 
Step 6) 1 is odd; subtract 1 and obtain 0.

Example 2:

Input: num = 8
Output: 4
Explanation: 
Step 1) 8 is even; divide by 2 and obtain 4. 
Step 2) 4 is even; divide by 2 and obtain 2. 
Step 3) 2 is even; divide by 2 and obtain 1. 
Step 4) 1 is odd; subtract 1 and obtain 0.

Example 3:

Input: num = 123
Output: 12

 

Constraints:

• 0 <= num <= 106

Solutions

Solution 1

class Solution:
  def numberOfSteps(self, num: int) -> int:
    ans = 0
    while num:
      if num & 1:
        num -= 1
      else:
        num >>= 1
      ans += 1
    return ans

class Solution {

  public int numberOfSteps(int num) {
    int ans = 0;
    while (num != 0) {
      num = (num & 1) == 1 ? num - 1 : num >> 1;
      ++ans;
    }
    return ans;
  }
}

class Solution {
public:
  int numberOfSteps(int num) {
    int ans = 0;
    while (num) {
      num = num & 1 ? num - 1 : num >> 1;
      ++ans;
    }
    return ans;
  }
};

func numberOfSteps(num int) int {
  ans := 0
  for num != 0 {
    if (num & 1) == 1 {
      num--
    } else {
      num >>= 1
    }
    ans++
  }
  return ans
}

function numberOfSteps(num: number): number {
  let ans = 0;
  while (num) {
    num = num & 1 ? num - 1 : num >>> 1;
    ans++;
  }
  return ans;
}

impl Solution {
  pub fn number_of_steps(mut num: i32) -> i32 {
    let mut count = 0;
    while num != 0 {
      if num % 2 == 0 {
        num >>= 1;
      } else {
        num -= 1;
      }
      count += 1;
    }
    count
  }
}

Solution 2

class Solution:
  def numberOfSteps(self, num: int) -> int:
    if num == 0:
      return 0
    return 1 + (
      self.numberOfSteps(num // 2)
      if num % 2 == 0
      else self.numberOfSteps(num - 1)
    )

class Solution {

  public int numberOfSteps(int num) {
    if (num == 0) {
      return 0;
    }
    return 1 + numberOfSteps((num & 1) == 0 ? num >> 1 : num - 1);
  }
}

class Solution {
public:
  int numberOfSteps(int num) {
    if (num == 0) return 0;
    return 1 + (num & 1 ? numberOfSteps(num - 1) : numberOfSteps(num >> 1));
  }
};

func numberOfSteps(num int) int {
  if num == 0 {
    return 0
  }
  if (num & 1) == 0 {
    return 1 + numberOfSteps(num>>1)
  }
  return 1 + numberOfSteps(num-1)
}

impl Solution {
  pub fn number_of_steps(mut num: i32) -> i32 {
    if num == 0 {
      0
    } else if num % 2 == 0 {
      1 + Solution::number_of_steps(num >> 1)
    } else {
      1 + Solution::number_of_steps(num - 1)
    }
  }
}