代码随想录算法训练营第23天 | 第七章 回溯算法part02

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今日记录

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39.组合总和

Leetcode链接

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

40.组合总和Ⅱ

Leetcode链接

难点：candidates里有重复元素，但是最后的集合不能重复
将candidates进行排序（sort）后循环时判决 i>index && candidates[i] == candidates[i - 1]

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

131.分割回文串

Leetcode链接

难点：

1. 判断回文串（isPalindrome—双指针法）
2. 分割和回溯的关系

class Solution {
private:
    vector<vector<string>> result;
    vector<string> path;
    void backtracking(string s, int index) {
        if (index >= s.size()) {
            result.push_back(path);
            return;
        }
        for (int i = index; i < s.size(); i++) {
            if (isPalindrome(s, index, i)) {
                string str = s.substr(index, i - index + 1);
                path.push_back(str);
            } else {
                continue;
            }
            backtracking(s, i + 1);
            path.pop_back();
        }
    }
    bool isPalindrome(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<vector<string>> partition(string s) {
        result.clear();
        path.clear();
        backtracking(s, 0);
        return result;
    }
};

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总结

原文地址：https://blog.csdn.net/monoki/article/details/140095861

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