Day.20 | 39.组合总和 40.组合总和II 131.分割回文串

39.组合总和

要点：元素可以重复，那么遍历的startIndex就跟之前的回溯算法不一样了；

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

    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backtracking(candidates, target, 0, 0);
        return result;
    }
};

40.组合总和II

要点：如何在同树层标记相等的元素已经使用过，加入了一个used状态数组，这里的使用还需要多理解

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

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<bool> used(candidates.size(), false);
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, target, 0, 0, used);
        return result;
    }
};

131.分割回文串

要点：双指针判断字符串是否是回文串。

class Solution {
public:
    bool isPalindrome(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;
    }

    vector<vector<string>> result;
    vector<string> path;
    void backtracking(string s, int startIndex) {
        if (startIndex >= s.size()) {
            result.push_back(path);
            return;
        }
        for (int i = startIndex; i < s.size(); i++) {
            if (isPalindrome(s, startIndex, i)) {
                string sub = s.substr(startIndex, i - startIndex + 1);
                path.push_back(sub);
            } else {
                continue;
            }
            backtracking(s, i + 1);
            path.pop_back();
        }
    }
    vector<vector<string>> partition(string s) {
        backtracking(s, 0);
        return result;
    }
};