leetcode39 组合总和

leetcode39 组合总和

思路：

这题也是用回溯，但是考虑到不能重复，所以用一个currentNum记录当前遍历到哪了，按照大小顺序遍历防止重复就行。

class Solution {//11：10-11：18public void dfs(int[] candidates,List<List<Integer>> ans,int currentNum,List<Integer> list,int sum,int target){if (sum==target){ans.add(new ArrayList<>(list));}if(sum>target){return;}if (currentNum>candidates.length){return;}for (int i = currentNum; i < candidates.length; i++) {int num=candidates[i];list.add(num);dfs(candidates,ans,i,list,sum+num,target);list.remove(list.size()-1);}}public List<List<Integer>> combinationSum(int[] candidates, int target) {List<List<Integer>> ans=new ArrayList<>();dfs(candidates,ans,0,new ArrayList<>(),0,target);return ans;}
}

如果不用currentNum的话，就是遍历所有的组合了（带顺序的）。

class Solution {//11：10-11：18public void dfs(int[] candidates,List<List<Integer>> ans,List<Integer> list,int sum,int target){if (sum==target){ans.add(new ArrayList<>(list));}if(sum>target){return;}for (int i = 0; i < candidates.length; i++) {int num=candidates[i];list.add(num);dfs(candidates,ans,list,sum+num,target);list.remove(list.size()-1);}}public List<List<Integer>> combinationSum(int[] candidates, int target) {List<List<Integer>> ans=new ArrayList<>();dfs(candidates,ans,new ArrayList<>(),0,target);return ans;}
}