标签：sum pre turn res for system continue tin nat
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] ]
class Solution { public List<List<Integer>> combinationSum2(int[] candidates, int target) { if(candidates.length==0) return new ArrayList(); Arrays.sort(candidates); List<List<Integer>> results = new ArrayList(); List<Integer> result = new ArrayList(); dfs(candidates, results, result, target, 0); return results; } public void dfs(int[] candidates,List<List<Integer>> results,List<Integer> result, int target, int start ){ if(target==0){ System.out.println(result); results.add(new ArrayList(result)); return; } int pre = -1;
for(int i = start; i < candidates.length; i++){ if(pre==candidates[i]) continue;//用previous防止重复 if((target-candidates[i]) < 0){ return; } pre = candidates[i]; result.add(candidates[i]); dfs(candidates, results, result, target-candidates[i],i+1);//i变成i+1 result.remove(result.size()-1); } } }
// 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]最多只用一次标签：sum pre turn res for system continue tin nat
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/10474436.html
