标签：直接 where arrays ret 思想 lis turn i++ generated
Descirption:
Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:
[ [7], [2, 2, 3] ]
Thoughts:
要解决这个问题要使用回溯的思想。所谓的回溯就是对所有的解进行枚举，能够走通的就继续走，否则就回退一步，尝试另一条路径，直到所有的路径都进行了遍历。
对于当前的回溯点我们需要直到当前所有可能的路径，然后朝着所有可能的路径继续前进，对于不可能的路径直接跳过
以下是java代码package middle;
import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class CombinationSum { public List<List<Integer>> combinationSum(int[] candidates, int target){ List<List<Integer>> list = new ArrayList<List<Integer>>(); Arrays.sort(candidates); backtrack(list, new ArrayList(),candidates, target, 0); /* System.out.println(list);*/ return list; } private void backtrack(List<List<Integer>> list, ArrayList arrayList, int[] candidates, int target, int start) { // TODO Auto-generated method stub if(target < 0){ return; }else if(target == 0){
//we should new an ArrayList, because the arrayList will change later,
//if we do not copy it‘s value, list will add []
list.add(new ArrayList<Integer>(arrayList)); }else{ for(int i = start;i<candidates.length;i++){ arrayList.add(candidates[i]); backtrack(list, arrayList, candidates, target-candidates[i], i); arrayList.remove(arrayList.size() - 1); } } } public static void main(String[] args){ CombinationSum sum = new CombinationSum(); int[] candidates = new int[]{2, 3,5 ,7}; sum.combinationSum(candidates, 7); } }
标签：直接 where arrays ret 思想 lis turn i++ generated
原文地址：http://www.cnblogs.com/whatyouknow123/p/7497155.html
