标签：rip and 元素 i++ get ons tar action arrays
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:
target) will be positive integers.Example 1:
Input: candidates =[10,1,2,7,6,1,5], target =8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]
题意：给一个数组，找出所有列表满足，列表和等于target
题不难，回溯算法，大家注意一点就行了，有重复的元素，使用set
class Solution { public List<List<Integer>> combinationSum2(int[] candidates, int target) { Set<List<Integer>> res = new HashSet<>(); if (candidates.length == 0) return new ArrayList<>(res); Arrays.sort(candidates); DFS(res, new ArrayList<Integer>(), candidates, target, 0, 0); return new ArrayList<>(res); } private void DFS(Set<List<Integer>> res, ArrayList<Integer> list, int[] candidates, int target, int index, int sum) { for (int i = index; i < candidates.length; i++) { sum += candidates[i]; list.add(candidates[i]); if (sum == target) res.add(new ArrayList<>(list)); else if (sum < target) DFS(res, list, candidates, target, i + 1, sum); else if (sum > target) { sum -= candidates[i]; list.remove(list.size() - 1); break; } sum -= candidates[i]; list.remove(list.size() - 1); } } }
[LeetCode] 40. Combination Sum II
标签：rip and 元素 i++ get ons tar action arrays
原文地址：https://www.cnblogs.com/Moriarty-cx/p/9942971.html
