Iterative and Recursion Algorithms to Compute the Number of Step

  • 时间:2020-09-11 08:17:29
  • 分类:网络文摘
  • 阅读:62 次

Given a non-negative integer num, return the number of steps to reduce it to zero. 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 <= 10^6

Hints:
Simulate the process to get the final answer.

Iterative Approach

We can simulate the process until we make the number zero (simple, intuitive and effective).

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
class Solution {
public:
    int numberOfSteps (int num) {
        int r = 0;
        while (num != 0) {
            r ++;
            if (num % 2 == 0) {
                num >>= 1;
            } else {
                num --;
            }
        }
        return r;
    }
};
class Solution {
public:
    int numberOfSteps (int num) {
        int r = 0;
        while (num != 0) {
            r ++;
            if (num % 2 == 0) {
                num >>= 1;
            } else {
                num --;
            }
        }
        return r;
    }
};

Recursive Algorithm

Alternatively, we can do this recursively but this approach is at the risk of stack-overflow.

1
2
3
4
5
6
7
class Solution {
public:
    int numberOfSteps (int num) {
        return num == 0 ? 0 : 1 + 
                numberOfSteps(num % 2 == 0 ? num / 2 : num - 1);
    }
};
class Solution {
public:
    int numberOfSteps (int num) {
        return num == 0 ? 0 : 1 + 
                numberOfSteps(num % 2 == 0 ? num / 2 : num - 1);
    }
};

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读:
苏秦以连横说秦原文及翻译  晋献文子成室原文及翻译  杜蒉扬觯原文及翻译  公子重耳对秦客原文及翻译  有子之言似夫子原文及翻译  曾子易箦原文及翻译  晋献公杀世子申生原文及翻译  关于描写春节习俗的作文  我要的效果  我喜欢的经典作品550字 
评论列表
添加评论