代码随想录算法训练营 | 回溯算法part02

39. 组合总和

cpp 复制代码

class Solution {
    vector<vector<int>> res;
    vector<int> temp;
public:
    void backtracing(vector<int>& candidates, int startIndex, int target, int sum){
        if (sum > target) {
            return;
        }
        if (sum == target) {
            res.push_back(temp);
            return;
        }
        //无重复元素 但是元素可以被多次使用
        //不能往回搜索 下一轮搜索的起点依然是 i
        for (int i = startIndex; i < candidates.size(); ++i) {
            sum += candidates[i];
            temp.push_back(candidates[i]);
            backtracing(candidates, i, target, sum);  //表示可以重复取当前的数
            sum -= candidates[i];
            temp.pop_back();
        }
        return ;
        
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backtracing(candidates, 0, target, 0);
        return res;
    }
};

40.组合总和II

cpp 复制代码

class Solution {
    vector<vector<int>> res;
    vector<int> temp;
public:
    void backtracing(vector<int>& candidates, int target, vector<bool>& used, int startIndex, int sum){
        if(sum == target) {
            res.push_back(temp);
            return;
        }
        for(int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; ++i) {
            // used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
            // used[i - 1] == false，说明同一树层candidates[i - 1]使用过
            // 要对同一树层使用过的元素进行跳过
            if( i > 0 && used[i - 1] == false && candidates[i] == candidates[i - 1]) {
                continue;
            }
            used[i] = true;
            sum += candidates[i];
            temp.push_back(candidates[i]);
            backtracing(candidates, target, used, i + 1, sum);
            temp.pop_back();
            sum -= candidates[i];
            used[i] = false;
        }
        return;
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end()); //需要排序，符合条件的组合也是有序的，便于去掉重复组合
        vector<bool> used(candidates.size(), false);
        backtracing(candidates, target, used, 0, 0);
        return res;
    }
};

131.分割回文串

cpp 复制代码

class Solution {
private:
    vector<vector<string>> res;
    vector<string> temp;
    void bactracing(const string& s, int startIndex) {
        if (startIndex >= s.size()) {
            res.push_back(temp);
            return;
        }
        for (int i = startIndex; i < s.size(); ++i) {
            if (isPalindrome(s, startIndex, i)) { // 是否是回文串
                string str = s.substr(startIndex, i - startIndex + 1); // 获取子串
                temp.push_back(str);
            } else {
                continue;
            }
            bactracing(s, i + 1);
            temp.pop_back();
        }
    }
    bool isPalindrome (const string& s, int left, int right) {
        for (int i = left, j = right; i < j; ++i, --j) {
            if (s[i] != s[j]) {
                return false;
            }
        }
        return true;
    }

public:
    vector<vector<string>> partition(string s) {
        bactracing(s, 0);
        return res;
    }
};