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:138 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:
数学题:甲、乙各存款若干元  倒推法练习题  数学题:一群猴子吃桃子  数学题:甲、乙、丙、丁四人共植树60棵  数学题:货车15小时走完全程  数学题:玉龙粮食加工厂生产一批面粉  数学题:从甲地到乙地的公路,只有上坡路和下坡路  数学题:求MN的长是多少  数学题:小红上学时坐车,回家步行  数学题:某班在植树活动中 
评论列表
添加评论