网站建设改代码改哪些nba最快的绝杀

思路

需要实现树枝层面的去重，利用use数组来判别，如果前一个节点已经使用了，说明这是在往深处遍历了，允许重复，如果前一个节点没有使用且值相同的话，说明是在树枝上重复了

代码

class Solution {
public:vector<int> path;vector<bool> used;vector< vector<int>> result;int curSum;void backtracking(vector<int>& candidates, int target, int startIndex){if(curSum > target) return;if(curSum == target){result.push_back(path);return;}for(int i = startIndex; i < candidates.size(); i++){if(i > 0 && candidates[i] == candidates[i-1] && !used[i-1]){continue;}used[i] = true;path.push_back(candidates[i]);curSum += candidates[i];backtracking(candidates, target, i+1);curSum -= candidates[i];path.pop_back();used[i] = false;}}vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {sort(candidates.begin(), candidates.end());for(int i = 0; i < candidates.size(); i++){used.push_back(false);}backtracking(candidates, target, 0);return result;}
};