[Leetcode] 130, 131, 62

130. Surrounded Regions

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,

X X X XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

Solution: 广度优先搜索(BFS)，最外面一圈的O是一定保留的，然后跟它们相连的O也能被保留下来。

Code:

class Solution {public:    void solve(vector<vector<char>>& board) {        if(board.size()==0 || board[0].size()==0) return;        queue<pair<int, int>> q;        //unordered_map<int, bool> visited; 这样做标记太浪费空间了，直接记录在board里面        //第一行        for(int i=0; i<board[0].size(); i++){            if(board[0][i] == 'O'){                q.push(make_pair(0, i));                board[0][i] = '+';//即能标记又能去重            }             if(board.back()[i] == 'O'){                q.push(make_pair(board.size()-1, i));                board[board.size()-1][i] = '+';            }        }        for(int i=0; i<board.size(); i++){            if(board[i][0] == 'O'){                q.push(make_pair(i, 0));                board[i][0] = '+';            }            if(board[i].back() == 'O'){                q.push(make_pair(i, board[0].size()-1));                board[i][board[0].size()-1] = '+';            }        }                while(!q.empty()){            int x = q.front().first;            int y = q.front().second;            if(x>0 && board[x-1][y]=='O'){                q.push(make_pair(x-1,y));                board[x-1][y] = '+';            }            if(y>0 && board[x][y-1]=='O'){                q.push(make_pair(x,y-1));                board[x][y-1] = '+';            }            if(x<board.size()-1 && board[x+1][y]=='O'){                q.push(make_pair(x+1,y));                board[x+1][y] = '+';            }            if(y<board[0].size()-1 && board[x][y+1]=='O'){                q.push(make_pair(x,y+1));                board[x][y+1] = '+';            }            q.pop();        }                for(int i=0; i<board.size(); i++){            for(int t=0; t<board[i].size(); t++){                if(board[i][t]=='O')                    board[i][t]='X';                else if(board[i][t]=='+')                    board[i][t]='O';            }        }    }};

131. Palindrome Partitioning

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"]]

Solution(1):深搜，每一步都将string分成两部分，一部分为一定包第一个字母的字符串，对剩下的字母构成的字符串继续递归，时间复杂度O(n^2)，空间复杂度O(n)。

Code:

class Solution {public:    vector<vector<string>> partition(string s) {        vector<vector<string>> ans;        vector<string> path;         dfs(0, s.size(), s, path, ans);        return ans;    }private:    bool isPalindrome(string& s){        for(int i=0, j=s.size()-1; i<j; i++,j--){            if(s[i]!=s[j]) return false;        }        return true;    }    void dfs(int first, int end, string& s, vector<string>& path, vector<vector<string>>& ans){        if(first == end){            ans.push_back(path);            return;        }        for(int i=1; i<=end-first; i++){            string sub = s.substr(first, i);            if(isPalindrome(sub)){                path.push_back(sub);                dfs(first+i, end, s, path, ans);                path.pop_back();            }        }    }};

Solution(2): 动态规划。

Code:

class Solution {public:    vector<vector<string>> partition(string s) {        int n = s.size();        if(n==0){            vector<vector<string>> ans;            ans.push_back(vector<string>());            return ans;        }        //预处理，也是动规思想        bool Palindrome[n][n];//Palindrome[x][y]，x-y之间的子串是否为回文        fill_n(&Palindrome[0][0], n*n, false);//Palindrome[x][y]要求x<=y，因此实际上真正有用的空间只有一半        for(int i=0; i<n; i++){            for(int t=i; t>=0; t--)                if(t<i-2 && i>0)                    Palindrome[t][i] = s[i]==s[t] && Palindrome[t+1][i-1];                else                    Palindrome[t][i] = s[i]==s[t];        }        //动规        vector<vector<string>> partition[n];//记录前n个字符partition的解集        for(int i=0; i<n; i++){//加入一个新字符            for(int t=i; t>0; t--){//遍历新字符作为最后一个能构成的所有子串                string sub = s.substr(t,i-t+1);                if(isPalindrome(sub)){                    //如果子串是回文，则将不包含子串在内的前t个字符的解集逐一加上这个回文加入当前寻找的解集                    for(vector<string> v:partition[t-1]){                        v.push_back(sub);                        partition[i].push_back(v);                    }                }            }            //t=0的情况            string sub = s.substr(0,i+1);            if(isPalindrome(sub)) partition[i].push_back(vector<string>{sub});        }        return partition[n-1];    }private:    bool isPalindrome(string& s){        for(int i=0, j=s.size()-1; i<j; i++,j--){            if(s[i]!=s[j]) return false;        }        return true;    }};

62. Unique Paths

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Solution(1): 深搜，直接深搜会超时，使用数组记录已经算过的值（备忘录法）。

Code:

class Solution {public:    int uniquePaths(int m, int n) {        //this->f = vector<vector<int> >(m + 1, vector<int>(n + 1, 0));         //会报runtime erro，free()的时候无法成功，为什么？        for(int i=0; i<=m; i++){            up.push_back(vector<int>());            for(int t=0; t<=n; t++){                up.back().push_back(0);            }        }        return dfs(m, n);    }private:    vector<vector<int>> up;    int dfs(int m, int n) {        if(m<=0 || n<=0) return 0;        if(m==1 || n==1) return 1;        if(up[m-1][n]==0) up[m-1][n] = dfs(m-1, n);        if(up[m][n-1]==0) up[m][n-1] = dfs(m, n-1);        return up[m-1][n]+up[m][n-1];    }};

Solution(2): 动规，设状态为up[i][j]，表示从起点(1; 1) 到达(i; j) 的路线条数，则状态转移方程为：up[i][j]=up[i-1][j]+up[i][j-1]

Code:

class Solution {public:    int uniquePaths(int m, int n) {        int up[m][n];        for(int i=0; i<n; i++) up[m-1][i] = 1;        for(int i=0; i<m; i++) up[i][n-1] = 1;        for(int i=m-2; i>=0; i--){            for(int t=n-2; t>=0; t--){                up[i][t] = up[i+1][t]+up[i][t+1];            }        }        return up[0][0];    }};

Solution(3):数学方法，机器人一共需要走m+n-2步，其中有m-1步是向下走，问一共有多少种走法，即m+n-2步中选择m-1步向下一共有多少种组合方式，即C(m+n-2,m-1)。

Code:

class Solution {public:    int uniquePaths(int m, int n) {        //C(m+n-2,m-1)        int k = min(m,n);//防止溢出        long long ans = 1;//防止溢出        for(int i=0,j=m+n-2; i<k-1; i++,j--)            ans *= j;        for(int i=1; i<=k-1; i++)            ans /= i;        return ans;    }};