397. Integer Replacement

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Problem

Given a positive integer n, you can apply one of the following operations:

1. If n is even, replace n with n / 2.
2. If n is odd, replace n with either n + 1 or n - 1.

Return the minimum number of operations needed for n to become 1.

Example 1:

Input: n = 8
Output: 3
Explanation: 8 -> 4 -> 2 -> 1

Example 2:

Input: n = 7
Output: 4
Explanation: 7 -> 8 -> 4 -> 2 -> 1
or 7 -> 6 -> 3 -> 2 -> 1

Example 3:

Input: n = 4
Output: 2

Code

class Solution {
    int res = Integer.MAX_VALUE;
    public int integerReplacement(int n) {
        helper((long)n, 0);
        return res;
    }

    private void helper(long n, int count) {
        if(n < 1) return;

        if(n == 1) {
            res = Math.min(res, count);
            return;
        }

        if ((n & 1) != 0) {
            helper(n + 1, count + 1);
            helper(n - 1, count + 1);
        } else {
            helper(n >> 1, count + 1);
        }
    }
}

当n为奇数的时, 除了3和7, +1之后是4的倍数的奇数, 应该+1, 比如15:

15 -> 16 -> 8 -> 4 -> 2 -> 1

15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1

7: +1和-1的结果相同

• 111 -> 110 -> 11 -> 10 -> 1
• 111 -> 1000 -> 100 -> 10 -> 1

3: -1

• 11 -> 10 -> 1
• 11 -> 100 -> 10 -> 1

class Solution {
    public int integerReplacement(int n) {
        if (n == Integer.MAX_VALUE) return 32; // n = 2^31-1
        int count = 0;
        while (n > 1){
            if (n % 2 == 0) {
                n /= 2;
            } else {
                // (n + 1) % 4 == 0
                if ((n & 2) != 0 && (n != 3)) {
                    n++;
                } else {
                    n--;
                }
            }
            count++;
        }
        return count;
    }
}