LeetCode 之 DFS 深度优先遍历

1. Palindrome Partioning II

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [    ["aa","b"],    ["a","a","b"]  ]

输出所有可能结果这样的问题，大部分用DFS

bool isPalindrome(string &s, int start, int end){    while(start < end){        if(s[start] != s[end])            return false;        start++, end--;    }    return true;}void DFSPalin(vector<vector<string>> &result,              vector<string> solu,              string &s,              int start){    if(start == s.size()){        result.push_back(solu);        return;    }    for(int i= start; i<s.size();i++){        if(isPalindrome(s,start,i)){            solu.push_back(s.substr(start,i-start+1));            DFSPalin(result,solu,s,i+1);            solu.pop_back();        }    }}vector<vector<string>> partition(string s) {    vector<vector<string>> result;    vector<string> solution;    DFSPalin(result,solution,s,0);    return result;}

2. Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

Analysis:

For the "Return all" problems, usually DFS or BFS will work well.
In this problem, a DFS with a simple cutting edge condition will pass all the test cases.

void dfsWordBreak(string &s,                  int st,                  vector<int>&sp,                  unordered_set<string> &dict,                  vector<string> &result,                  vector<vector<bool>> &mp){    if(st>=s.size()){        string str = s;        for(int i=0;i<sp.size()-1;i++){            str.insert(sp[i]+i," ");        }        result.push_back(str);    }else{        for(int j=0;j<mp[st].size();j++){            if(mp[st][j]==true){                sp.push_back(j+1);                dfsWordBreak(s,j+1,sp,dict,result,mp);                sp.pop_back();            }        }    }}vector<string> wordBreak2(string s, unordered_set<string> &dict) {    vector<string> result;    vector<vector<bool>> mp(s.size(),vector<bool>(s.size(),false));    for(int i=0;i<s.size();i++){        for(int j=i;j<s.size();j++){            if(dict.find(s.substr(i,j-i+1))!=dict.end())                mp[i][j] = true;        }    }    bool flag = false;    for(int i=0;i<s.size();i++){        if(mp[i][s.size()-1])            {flag = true;            break;}    }    if(!flag) return result;    vector<int>sp;    dfsWordBreak(s,0,sp,dict,result,mp);    return result;}