|
| 1 | +/** |
| 2 | + * [39] 组合总和 |
| 3 | + * |
| 4 | + * 给你一个 无重复元素 的整数数组 candidates 和一个目标整数 target ,找出 candidates 中可以使数字和为目标数 target 的 所有 不同组合 ,并以列表形式返回。你可以按 任意顺序 返回这些组合。 |
| 5 | + * candidates 中的 同一个 数字可以 无限制重复被选取 。如果至少一个数字的被选数量不同,则两种组合是不同的。 |
| 6 | + * 对于给定的输入,保证和为 target 的不同组合数少于 150 个。 |
| 7 | + * |
| 8 | + * 示例 1: |
| 9 | + * |
| 10 | + * 输入:candidates = [2,3,6,7], target = 7 |
| 11 | + * 输出:[[2,2,3],[7]] |
| 12 | + * 解释: |
| 13 | + * 2 和 3 可以形成一组候选,2 + 2 + 3 = 7 。注意 2 可以使用多次。 |
| 14 | + * 7 也是一个候选, 7 = 7 。 |
| 15 | + * 仅有这两种组合。 |
| 16 | + * 示例 2: |
| 17 | + * |
| 18 | + * 输入: candidates = [2,3,5], target = 8 |
| 19 | + * 输出: [[2,2,2,2],[2,3,3],[3,5]] |
| 20 | + * 示例 3: |
| 21 | + * |
| 22 | + * 输入: candidates = [2], target = 1 |
| 23 | + * 输出: [] |
| 24 | + * |
| 25 | + * |
| 26 | + * 提示: |
| 27 | + * |
| 28 | + * 1 <= candidates.length <= 30 |
| 29 | + * 2 <= candidates[i] <= 40 |
| 30 | + * candidates 的所有元素 互不相同 |
| 31 | + * 1 <= target <= 40 |
| 32 | + * |
| 33 | + */ |
| 34 | +pub struct Solution {} |
| 35 | + |
| 36 | +// problem: https://leetcode.cn/problems/combination-sum/ |
| 37 | +// discuss: https://leetcode.cn/problems/combination-sum/discuss/?currentPage=1&orderBy=most_votes&query= |
| 38 | + |
| 39 | +// submission codes start here |
| 40 | + |
| 41 | +impl Solution { |
| 42 | + pub fn combination_sum(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> { |
| 43 | + let mut candidates = candidates.clone(); |
| 44 | + candidates.sort(); |
| 45 | + |
| 46 | + let mut res = vec![]; |
| 47 | + |
| 48 | + let mut curr = vec![]; |
| 49 | + let mut curr_idx = vec![]; |
| 50 | + let mut sum = 0; |
| 51 | + |
| 52 | + curr.push(candidates[0]); |
| 53 | + curr_idx.push(0); |
| 54 | + sum += candidates[0]; |
| 55 | + |
| 56 | + while curr.len() > 0 { |
| 57 | + // println!("{:?} {} {}", curr, sum, target); |
| 58 | + if sum < target { |
| 59 | + let aidx = curr_idx[curr_idx.len()-1]; |
| 60 | + curr.push(candidates[aidx]); |
| 61 | + curr_idx.push(aidx); |
| 62 | + sum += candidates[aidx]; |
| 63 | + continue; |
| 64 | + } |
| 65 | + if sum == target { |
| 66 | + res.push(curr.clone()); |
| 67 | + } |
| 68 | + while let Some(i) = curr_idx.pop() { |
| 69 | + let v = curr.pop().unwrap(); |
| 70 | + sum -= v; |
| 71 | + |
| 72 | + if i + 1 < candidates.len() { |
| 73 | + let i: usize = i + 1; |
| 74 | + curr.push(candidates[i]); |
| 75 | + curr_idx.push(i); |
| 76 | + sum += candidates[i]; |
| 77 | + break; |
| 78 | + } |
| 79 | + } |
| 80 | + } |
| 81 | + res |
| 82 | + } |
| 83 | +} |
| 84 | + |
| 85 | +// submission codes end |
| 86 | + |
| 87 | +#[cfg(test)] |
| 88 | +mod tests { |
| 89 | + use super::*; |
| 90 | + |
| 91 | + #[test] |
| 92 | + fn test_39() { |
| 93 | + let empty_res: Vec<Vec<i32>> = vec![]; |
| 94 | + |
| 95 | + assert_eq!( |
| 96 | + Solution::combination_sum(vec![2, 3, 6, 7], 7), |
| 97 | + vec![vec![2, 2, 3], vec![7]] |
| 98 | + ); |
| 99 | + assert_eq!(Solution::combination_sum(vec![2], 1), empty_res); |
| 100 | + assert_eq!( |
| 101 | + Solution::combination_sum(vec![2, 3, 5], 8), |
| 102 | + vec![vec![2, 2, 2, 2], vec![2, 3, 3], vec![3, 5]] |
| 103 | + ); |
| 104 | + } |
| 105 | +} |
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