《慕课网玩转算法面试》笔记及习题解答8.1~8.3

LeetCode 17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

思路：采用回溯法，当当前是string长度等于digits长度时，则找到了一个满足条件的string

class Solution {private:    const string record[10] = {" ","" ,"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};        void findCombinations(const string& digits, const int& index, string s, vector<string>& res){        //找到了一个满足条件的s        if( index == digits.length() ){            res.push_back(s);            return;        }        char c = digits[index];        string letters = record[c - '0'];        for(int i = 0; i < letters.length(); i++){            findCombinations( digits, index + 1, s + letters[i] , res );        }        return;    }public:        vector<string> letterCombinations(string digits) {        vector<string> res;        if(digits =="")            return res;                findCombinations(digits, 0, "", res);        return res;    }};

LeetCode 93

有点难，暂时不会做。。

LeetCode 131

思路：使用回溯法，对于每个能构成回文的字串放入数组，后半部分递归，递归结束后进行回溯

class Solution {private:    bool isPalindrome(string s)    {        int i = 0, j = s.length()-1;        while(i < j){            if(s[i]!=s[j])                return false;            i++;            j--;        }        return true;    }        void findPartition(const string& s, const int& start , vector<string>& p, vector<vector<string>>& res){        if(start == s.length()){            res.push_back(p);            return;        }        for(int i = start+1; i <= s.length(); i++){            //如果s[start,i)是回文，将s[start,i)放入数组中，后半部分递归            if( isPalindrome(s.substr(start,i-start)) ){                p.push_back(s.substr(start,i-start));                findPartition(s, i, p, res);                //注意递归结束后要进行回溯操作                p.pop_back();            }        }        return;    }public:    vector<vector<string>> partition(string s) {        vector<vector<string>> res;        if(s.empty())            return res;        vector<string> p;        findPartition(s, 0, p,res);        return res;    }};

LeetCode 46

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[  [1,2,3],  [1,3,2],  [2,1,3],  [2,3,1],  [3,1,2],  [3,2,1]]

思路：回溯法寻找，设置一个bool数组来判断是否包含，注意终止条件，递归结束后需要将bool数组修改来保证回溯

class Solution {private:    vector<bool> isUsed;        void findPermute(const vector<int>& nums, int index, vector<int>& num, vector<vector<int>>& res)    {        //找到一个满足条件的数组的终止条件        if(num.size() == nums.size())        {            res.push_back(num);            return;        }        for(int i = 0; i< nums.size(); i++){            if(!isUsed[i]){                num.push_back(nums[i]);                isUsed[i] = true;                findPermute(nums, index+1, num, res);                //注意，递归结束时是已经找到了一个包含nums[index]的数组，要重新寻找新数组，此时要将nums[i] pop并且将isUesd[i]设置成false                num.pop_back();                isUsed[i] = false;                            }        }        return;            }public:    vector<vector<int>> permute(vector<int>& nums) {        vector<vector<int>> res;        if(nums.empty())            return res;        isUsed = vector<bool>(nums.size(),false);        vector<int> num;        findPermute(nums, 0, num, res);        return res;    }};