【代码随想录】day27

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一、39组合总和

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int>& nums, int sumNum, int target, int startNum) {
        if (sumNum == target) {
            res.push_back(path);
        }
        if (sumNum >= target || startNum == nums.size()) {
            return;
        }
        for (int i = startNum; i < nums.size(); i ++) {
            path.push_back(nums[i]);
            backtracking(nums, sumNum + nums[i], target, i);
            path.pop_back();
        }

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

二、40组合总和II

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int>& nums, int sum, int target, int startNum) {
        if (sum == target) {
            res.push_back(path);
        }
        if (sum >= target || startNum == nums.size()) {
            return;
        }
        for (int i = startNum; i < nums.size(); i ++) {
            if (path.size() == 0 && i > 0 && nums[i] == nums[i-1]) {
                continue;
            }
            if (i > startNum && nums[i] == nums[i-1]) {
                continue;
            }
            path.push_back(nums[i]);
            backtracking(nums, sum + nums[i], target, i + 1);
            path.pop_back();
        }

    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, 0, target, 0);
        return res;
    }
};

使用used数组树层去重很难想，感觉不如直接画出树形图按广度、深度分别写过滤条件来得容易。

三、131分割回文串

class Solution {
public:
    vector<vector<string>> res;
    vector<string> path;
    bool isValid(string s, int startIndex, int endIndex) {
        for (int left = startIndex, right = endIndex - 1; left < right; left ++, right --) {
            if (s[left] != s[right]) {
                return false;
            }
        }
        return true;
    }

    void backtracking(string s, int length) {
        if (length == s.size()) {
            res.push_back(path);
        }
        if (length >= s.size()) {
            return;
        }
        for (int len = 1; len <= s.size() - length; len ++) {
            if (isValid(s, length, length + len) == false) {
                continue;
            }
            string temp = s.substr(length, len);
            path.push_back(temp);
            backtracking(s, length + temp.size());
            path.pop_back();
        }
    }

    vector<vector<string>> partition(string s) {
        backtracking(s, 0);
        return res;
    }
};

总结

回溯法的题目主要是要先画出树形图，然后一步一步分析即可。