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Combination Sum II

描述

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1,a2,...,aka_1, a_2, ..., a_k) must be in non-descending order. (ie, a1>a2>...>aka_1 > a_2 > ... > a_k).
  • The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, A solution set is:

[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

分析

代码

// Combination Sum II
// 时间复杂度O(n!),空间复杂度O(n)
public class Solution {
public List<List<Integer>> combinationSum2(int[] nums, int target) {
Arrays.sort(nums); // 跟第 50 行配合,
// 确保每个元素最多只用一次
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
dfs(nums, path, result, target, 0);
return result;
}
// 使用nums[start, nums.size())之间的元素,能找到的所有可行解
private static void dfs(int[] nums, List<Integer> path,
List<List<Integer>> result, int gap, int start) {
if (gap == 0) { // 找到一个合法解
result.add(new ArrayList<>(path));
return;
}

int previous = -1;
for (int i = start; i < nums.length; i++) {
// 如果上一轮循环已经使用了nums[i],则本次循环就不能再选nums[i],
// 确保nums[i]最多只用一次
if (previous == nums[i]) continue;

if (gap < nums[i]) return; // 剪枝

previous = nums[i];

path.add(nums[i]);
dfs(nums, path, result, gap - nums[i], i + 1);
path.remove(path.size() - 1); // 恢复环境
}
}
}

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