LeetCode每日一题(397. Integer Replacement)

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

If n is even, replace n with n / 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

Constraints:

• 1 <= n <= 231 - 1

---

dp[n]是 n 变到 1 的最小步数， 如果 n 是偶数 dp[n] = dp[n/2] + 1, 如果 n 是奇数 dp[n] = min(dp[n-1], dp[n+1]), 注意， 题目本身 1 <= n <= 2.pow(32)-1, 但是如果考虑+1 的情况还是会存在 i32 溢出的问题， 所以还是要用 i64 或者 u32 来处理。

---

use std::collections::HashMap;

impl Solution {
    fn dp(n: i64, cache: &mut HashMap<i64, i32>) -> i32 {
        if n == 1 {
            return 0;
        }
        if n % 2 == 0 {
            let next = if let Some(c) = cache.get(&(n / 2)) {
                *c
            } else {
                Solution::dp(n / 2, cache)
            };
            cache.insert(n, next + 1);
            return next + 1;
        }
        let add = Solution::dp(n + 1, cache);
        let sub = Solution::dp(n - 1, cache);
        cache.insert(n, add.min(sub) + 1);
        add.min(sub) + 1
    }
    pub fn integer_replacement(n: i32) -> i32 {
        Solution::dp(n as i64, &mut HashMap::new())
    }
}