二刷代码随想录第23天

39. 组合总和

• 因为可以重复，传入start_index的值不需要+1了

class Solution {
public:
    vector<int> path;
    vector<vector<int>> result;
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backtracking(candidates, target, 0);
        return result;
    }

    void backtracking(vector<int>& candidates, int target, int start_idnex) {
        int sum = 0;
        for (auto n : path) {
            sum += n;
        }
        if (sum > target) {
            return;
        }
        if (sum == target) {
            result.push_back(path);
            return;
        }

        for (int i = start_idnex; i < candidates.size(); i++) {
            path.push_back(candidates[i]);
            backtracking(candidates, target, i);
            path.pop_back();
        }
    }
};

40. 组合总和 II

• 需要用used数组来确认相同的数字有没有被用过
• used[i - 1] == 0代表这一层中这个数没被取过

class Solution {
public:
    vector<int> path;
    vector<vector<int>> result;
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<int> used(candidates.size(), 0);
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, target, 0, used);
        return result;
    }

    void backtracking(vector<int>& candidates, int target, int start_index,
                      vector<int> used) {
        int sum = 0;
        for (int n : path) {
            sum += n;
        }
        if (sum > target) {
            return;
        }
        if (sum == target) {
            result.push_back(path);
            return;
        }

        for (int i = start_index;
             i < candidates.size() && sum + candidates[i] <= target; i++) {
            if (i > 0 && candidates[i] == candidates[i - 1] &&
                used[i - 1] == 0) {
                continue;
            }
            used[i] = 1;
            path.push_back(candidates[i]);
            backtracking(candidates, target, i + 1, used);
            path.pop_back();
            used[i] = 0;
        }
    }
};

131. 分割回文串

• statr_index是回文串开始位置，i是回文串结束位置

class Solution {
public:
    vector<string> path;
    vector<vector<string>> result;
    vector<vector<string>> partition(string s) {
        backtracking(s, 0);
        return result;
    }

    void backtracking(string s, int start_index) {
        if (start_index >= s.size()) {
            result.push_back(path);
            return;
        }
        for (int i = start_index; i < s.size(); i++) {
            if (is_back_word(s, start_index, i)) {
                string str = s.substr(start_index, i - start_index + 1);
                path.push_back(str);
                backtracking(s, i + 1);
                path.pop_back();
            }
        }
    }

    bool is_back_word(string s, int start_index, int end_index) {
        while (start_index < end_index) {
            if (s[start_index] != s[end_index]) {
                return false;
            }
            start_index++;
            end_index--;
        }
        return true;
    }
};