Combination Sum

来源：互联网发布：流程优化与再造编辑：程序博客网时间：2024/05/01 03:07

class Solution {public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        sort(candidates.begin(), candidates.end());        vector<vector<int> > result;        vector<int> path;        int sum = 0;        dfs(candidates, 0, sum, path, result, target);        return result;    }    void dfs(vector<int> &candidates, int pos, int &sum, vector<int> &path, vector<vector<int> > &result, int target)    {        if(sum == target)        {            result.push_back(path);            return;        }        else if(sum > target)            return;        else        {            for(int i=pos; i<candidates.size(); ++i)            {                path.push_back(candidates[i]);                sum += candidates[i];                dfs(candidates, i, sum, path, result, target);                sum -= candidates[i];                path.pop_back();            }        }    }};

9.06非递归实现：参考 http://blog.csdn.net/starmsg/article/details/39085143

注意：如果每次可以push相同的元素入栈，则无法通过

s.peek() == currentPath.peek()

区分是否有新的元素push到stack中，此时可以在path中记录stack top的index而非实际的元素，来标记是否有新的元素push。

新的判断条件变成了

<span style="color:#333333;">(stack.size()-1) == path.back()</span>

class Solution {public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        vector<vector<int> > result;        if(candidates.size() == 0)            return result;        sort(candidates.begin(), candidates.end());                for(int start = 0; start < candidates.size(); start++)            dfsWithStack(candidates, result, start, target);        return result;    }        void dfsWithStack(vector<int> &candidates, vector<vector<int> > &result, int startIdx, int target)    {        vector<int> path;        vector<int> stack;        int sum = 0;        if(sum + candidates[startIdx] <= target)        {            stack.push_back(startIdx);        }            int count = 0;            while(!stack.empty())        {            int idx = stack.back();            path.push_back(stack.size()-1);            sum += candidates[idx];                if(sum == target)            {            vector<int> res;            for(int i=0; i<path.size(); i++)            res.push_back(candidates[stack[path[i]]]);                result.push_back(res);            }            else if(sum < target)            {                for(size_t i = idx; i < candidates.size(); i++)                {                    if(sum + candidates[i] <= target)                    {                        stack.push_back(i);                    }                }            }                while(!stack.empty() && (stack.size()-1) == path.back())            {                sum -= candidates[stack.back()];                path.pop_back();                stack.pop_back();            }            count++;        }    }};

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