代码随想录算法训练营第27天|39. 组合总和 40.组合总和II 131.分割回文串

39. 组合总和

class Solution {
private:
    vector> res;
    vector path;
    void backtracking(vector& candidates, int target, int sum, int startIndex) {
        if (sum > target) return;
        if (sum == target) {
            res.push_back(path);
            return;
        }
        for (int i = startIndex; i < candidates.size(); i++) {
            path.push_back(candidates[i]);
            backtracking(candidates, target, sum + candidates[i], i);
            path.pop_back();
        }
    }
public:
    vector> combinationSum(vector& candidates, int target) {
        res.clear();
        path.clear();
        backtracking(candidates, target, 0, 0);
        return res;
    }
};

不用在for中backtracking部分传入i + 1了，这样才能重复使用元素。 

40.组合总和II

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

used数组的作用：区分树层和树枝去重，在树枝内used是true，不会因为与上一个相同就排除掉结果；树层内used是false，会因为与上一个相同直接continue。 

去重的基础：排序。

这里不使用used数组也是可以的，只需判断i是否大于startIndex。 

不使用sum也可以，通过target==0来判断。

131.分割回文串

class Solution {
private:
    vector path;
    vector> res;
    void backtracking(const string& s, int startIndex) {
        if (startIndex >= s.size()) {
            res.push_back(path);
            return;
        }
        for (int i = startIndex; i < s.size(); i++) {
            if (isP(s, startIndex, i)) {
                string str = s.substr(startIndex, i - startIndex + 1);
                path.push_back(str);
            }else {
                continue;
            }
            backtracking(s, i + 1);
            path.pop_back();
        }
    }
    bool isP(const string& s, int start, int end) {
        for (int i = start, j = end; i < j; i++, j--) {
            if (s[i] != s[j]) return false;
        }
        return true;
    }
public:
    vector> partition(string s) {
        path.clear();
        res.clear();
        backtracking(s, 0);
        return res;
    }
};