LeetCode每日一题(216. Combination Sum III)

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used.
Each number is used at most once.
Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]

Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]

Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.

Example 3:

Input: k = 4, n = 1
Output: []

Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Constraints:

• 2 <= k <= 9
• 1 <= n <= 60

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以前用 golang 实现过一次， 这次换 rust 来实现， 结果被一个小失误绊了一个多小时
用 dp 来解， dp[k][d][curr]代表还剩余 k 次挑选机会，与目标数 n 的差异为 d，当前数字为 curr 的情况下的所有组合， 要注意的是， 我们的 curr 一定得是单向递增的， 这样才不会出现挑选重复数字的情况。但是要注意的是， 最终返回边界条件是 k1 且 dcurr, 最开始我认为只要 d >= curr 且 d < 9 就可以返回， 但是这样会导致出现重复的组合。

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use std::collections::HashMap;

impl Solution {
    fn dp(k: i32, n: i32, curr: i32, cache: &mut HashMap<(i32, i32, i32), Vec<Vec<i32>>>) -> Vec<Vec<i32>> {
        let mut ans = Vec::new();
        if k == 1 {
            if n == curr {
                return vec![vec![n]];
            }
            return ans;
        }
        for i in curr + 1..10 {
            let mut next = if let Some(c) = cache.get(&(k - 1, n - curr, i)) {
                c.clone()
            } else {
                Solution::dp(k - 1, n - curr, i, cache)
            };
            next.iter_mut().for_each(|l| l.push(curr));
            ans.extend(next);
        }
        cache.insert((k, n, curr), ans.clone());
        ans
    }
    pub fn combination_sum3(k: i32, n: i32) -> Vec<Vec<i32>> {
        let mut ans = Vec::new();
        let mut cache = HashMap::new();
        for i in 1..10 {
            let next = Solution::dp(k, n, i, &mut cache);
            ans.extend(next);
        }
        ans
    }
}