LeetCode 2016 37,65,212,84,130,218

37 Sudoku Solver

// ----------------Seeing Discuss-----------
/*
	copying codes
	url
	https://discuss.leetcode.com/topic/5002/a-simple-dfs-solution

	Analysis:
		
*/
class Solution {
public:
	bool isValidSudoku(vector > &board) {
		return true;
	}
	void solveSudoku(vector > &board) {
		util(board, 0);
	}
	bool util(vector>& board, int pos)
	{
		if (pos >= 81)
			return true;
		int i = pos / 9;
		int j = pos % 9;
		if (board[i][j] != '.')
			return util(board, pos + 1);
		else
		{
			for (char c = '1'; c <= '9'; c++)
			{
				if (!isInRow(board, i,c) && !isInCol(board, j, c) && !isInRec(board, i, j, c))
				{
					board[i][j] = c;
					if (util(board, pos + 1))
						return true;
					else
						board[i][j] = '.';
				}
			}
			return false;
		}
	}

	bool isInRow(vector>& board, int i, char c)
	{
		vector& row = board[i];
		for (int k = 0; k < 9; k++)
		{
			if (row[k] == c)
				return true;
		}
		return false;
	}
	bool isInCol(vector>& board,int j, char c)
	{
		for (int k = 0; k < 9; k++)
		{
			if (board[k][j] == c)
				return true;
		}
		return false;
	}
	bool isInRec(vector>& board, int i, int j, char c)
	{
		int bigrow = i / 3, bigcol = j / 3;
		for (int m = 3 * bigrow; m < 3 * (bigrow + 1); m++)
		{
			for (int n = 3 * bigcol; n < 3 * (bigcol + 1); n++)
				if (board[m][n] == c)
					return true;
		}
		return false;
	}
};

65 Valid Number

// ----------------Seeing Discuss-----------
/*
	copying codes
	url
	https://discuss.leetcode.com/topic/2710/a-simple-solution-in-cpp
*/
class Solution {
public:
    bool isNumber(string s)
    {
        int i=0;
        for(;s[i]==' ';i++) {}
        if(s[i]=='+' || s[i]=='-') i++;
        int n_nm,n_pt;
        for(n_nm=0,n_pt=0;(s[i]<='9' && s[i]>='0')|| s[i]=='.';i++)
            s[i] =='.'?n_pt++:n_nm++;
        if(n_pt>1 || n_nm<1) return false;
        if(s[i]=='e')
        {
            i++;
            if(s[i]=='+' || s[i]=='-')i++;
            int n_nm=0;
            for(;s[i]>='0' && s[i]<='9';i++,n_nm++){}
            if(n_nm<1) return false;
        }
        for(;s[i]==' ';i++){}
        return s[i]==0;
    }
};

212 Word Search II

// ---------------------Seeing Discuss -------------
/*
	copying codes
	url:
		https://discuss.leetcode.com/topic/20210/my-ac-very-clean-c-code
	The idea is start from every position of the board, and then see if we can 
	find a word starting from this position with checking if is_end is true in TrieNode structure
	
*/
class TrieNode{
public:
    bool is_end;
    vector children;
    TrieNode(){
        is_end=false;
        children=vector(26, NULL);
    }   
};

class Trie{
public:
    TrieNode* getRoot(){return root;}
    Trie(vector& words){
        root=new TrieNode();
        for(int i=0; ichildren[index]==NULL)   
               cur->children[index]=new TrieNode();
            cur=cur->children[index];    
        }
        cur->is_end=true;
    }
private:
    TrieNode* root;
};

class Solution {
public:
    vector findWords(vector>& board, vector& words) {
        Trie* trie = new Trie(words);
        TrieNode* root=trie->getRoot();
        set result_set;
        for(int x=0; x result;
        for(auto it:result_set)    result.push_back(it);
        return result;        
    }
private:
    void findWords(vector>& board, int x, int y, TrieNode* root, string word, set& result){
        if(x<0||x>=board.size()||y<0||y>=board[0].size() || board[x][y]==' ') return;
        
        if(root->children[board[x][y]-'a'] != NULL){
            word=word+board[x][y];
            root=root->children[board[x][y]-'a']; 
            if(root->is_end) result.insert(word);
            char c=board[x][y];
            board[x][y]=' ';
            findWords(board, x+1, y, root, word, result);
            findWords(board, x-1, y, root, word, result);
            findWords(board, x, y+1, root, word, result);
            findWords(board, x, y-1, root, word, result);
            board[x][y]=c;        
        }
    }
};
// -------------------TLE 1st depth first search--------------------
// 35 / 37 test cases passed.
// the TLE data point:
// 		["aaaa","aaaa","aaaa","aaaa","bcde","fghi","jklm","nopq","rstu","vwxy","zzzz"]
// 		["aaaaaaaaaaaaaaaa","aaaaaaaaaaaaaaab","aaaaaaaaaaaaaaac","aaaaaaaaaaaaaaad","aaaaaaaaaaaaaaae","aaaaaaaaaaaaaaaf","aaaaaaaaaaaaaaag","aaaaaaaaaaaaaaah","aaaaaaaaaaaaaaai","aaaaaaaaaaaaaaaj","aaaaaaaaaaaaaaak","aaaaaaaaaaaaaaal","aaaaaaaaaaaaaaam","aaaaaaaaaaaaaaan","aaaaaaaaaaaaaaao","aaaaaaaaaaaaaaap","aaaaaaaaaaaaaaaq","aaaaaaaaaaaaaaar","aaaaaaaaaaaaaaas","aaaaaaaaaaaaaaat","aaaaaaaaaaaaaaau","aaaaaaaaaaaaaaav","aaaaaaaaaaaaaaaw","aaaaaaaaaaaaaaax","aaaaaaaaaaaaaaay","aaaaaaaaaaaaaaaz","aaaaaaaaaaaaaaaa","aaaaaaaaaaaaaaab","aaaaaaaaaaaaaaac","aaaaaaaaaaaaaaad","aaaaaaaaaaaaaaae","aaaaaaaaaaaaaaaf","aaaaaaaaaaaaaaag","aaaaaaaaaaaaaaah","aaaaaaaaaaaaaaai","aaaaaaaaaaaaaaaj","aaaaaaaaaaaaaaak","aaaaaaaaaaaaaaal","aaaaaaaaaaaaaaam","aaaaaaaaaaaaaaan","aaaaaaaaaaaaaaao","aaaaaaaaaaaaaaap","aaaaaaaaaaaaaaaq","aaaaaaaaaaaaaaar","aaaaaaaaaaaaaaas","aaaaaaaaaaaaaaat","aaaaaaaaaaaaaaau","aaaaaaaaaaaaaaav","aaaaaaaaaaaaaaaw","aaaaaaaaaaaaaaax","aaaaaaaaaaaaaaay","aaaaaaaaaaaaaaaz","aaaaaaaaaaaaaaba","aaaaaaaaaaaaaabb","aaaaaaaaaaaaaabc","aaaaaaaaaaaaaabd","aaaaaaaaaaaaaabe","aaaaaaaaaaaaaabf","aaaaaaaaaaaaaabg","aaaaaaaaaaaaaabh","aaaaaaaaaaaaaabi","aaaaaaaaaaaaaabj","aaaaaaaaaaaaaabk","aaaaaaaaaaaaaabl","aaaaaaaaaaaaaabm","aaaaaaaaaaaaaabn","aaaaaaaaaaaaaabo","aaaaaaaaaaaaaabp","aaaaaaaaaaaaaabq","aaaaaaaaaaaaaabr","aaaaaaaaaaaaaabs","aaaaaaaaaaaaaabt","aaaaaaaaaaaaaabu","aaaaaaaaaaaaaabv","aaaaaaaaaaaaaabw","aaaaaaaaaaaaaabx","aaaaaaaaaaaaaaby","aaaaaaaaaaaaaabz","aaaaaaaaaaaaaaca","aaaaaaaaaaaaaacb","aaaaaaaaaaaaaacc","aaaaaaaaaaaaaacd","aaaaaaaaaaaaaace","aaaaaaaaaaaaaacf","aaaaaaaaaaaaaacg","aaaaaaaaaaaaaach","aaaaaaaaaaaaaaci","aaaaaaaaaaaaaacj","aaaaaaaaaaaaaack","aaaaaaaaaaaaaacl","aaaaaaaaaaaaaacm","aaaaaaaaaaaaaacn","aaaaaaaaaaaaaaco","aaaaaaaaaaaaaacp","aaaaaaaaaaaaaacq","aaaaaaaaaaaaaacr","aaaaaaaaaaaaaacs","aaaaaaaaaaaaaact","aaaaaaaaaaaaaacu","aaaaaaaaaaaaaacv","aaaaaaaaaaaaaacw","aaaaaaaaaaaaaacx","aaaaaaaaaaaaaacy","aaaaaaaaaaaaaacz","aaaaaaaaaaaaaada","aaaaaaaaaaaaaadb","aaaaaaaaaaaaaadc","aaaaaaaaaaaaaadd","aaaaaaaaaaaaaade","aaaaaaaaaaaaaadf","aaaaaaaaaaaaaadg","aaaaaaaaaaaaaadh","aaaaaaaaaaaaaadi","aaaaaaaaaaaaaadj","aaaaaaaaaaaaaadk","aaaaaaaaaaaaaadl","aaaaaaaaaaaaaadm","aaaaaaaaaaaaaadn","aaaaaaaaaaaaaado","aaaaaaaaaaaaaadp","aaaaaaaaaaaaaadq","aaaaaaaaaaaaaadr","aaaaaaaaaaaaaads","aaaaaaaaaaaaaadt","aaaaaaaaaaaaaadu","aaaaaaaaaaaaaadv","aaaaaaaaaaaaaadw","aaaaaaaaaaaaaadx","aaaaaaaaaaaaaady","aaaaaaaaaaaaaadz","aaaaaaaaaaaaaaea","aaaaaaaaaaaaaaeb","aaaaaaaaaaaaaaec","aaaaaaaaaaaaaaed","aaaaaaaaaaaaaaee","aaaaaaaaaaaaaaef","aaaaaaaaaaaaaaeg","aaaaaaaaaaaaaaeh","aaaaaaaaaaaaaaei","aaaaaaaaaaaaaaej","aaaaaaaaaaaaaaek","aaaaaaaaaaaaaael","aaaaaaaaaaaaaaem","aaaaaaaaaaaaaaen","aaaaaaaaaaaaaaeo","aaaaaaaaaaaaaaep","aaaaaaaaaaaaaaeq","aaaaaaaaaaaaaaer","aaaaaaaaaaaaaaes","aaaaaaaaaaaaaaet","aaaaaaaaaaaaaaeu","aaaaaaaaaaaaaaev","aaaaaaaaaaaaaaew","aaaaaaaaaaaaaaex","aaaaaaaaaaaaaaey","aaaaaaaaaaaaaaez","aaaaaaaaaaaaaafa","aaaaaaaaaaaaaafb","aaaaaaaaaaaaaafc","aaaaaaaaaaaaaafd","aaaaaaaaaaaaaafe","aaaaaaaaaaaaaaff","aaaaaaaaaaaaaafg","aaaaaaaaaaaaaafh","aaaaaaaaaaaaaafi","aaaaaaaaaaaaaafj","aaaaaaaaaaaaaafk","aaaaaaaaaaaaaafl","aaaaaaaaaaaaaafm","aaaaaaaaaaaaaafn","aaaaaaaaaaaaaafo","aaaaaaaaaaaaaafp","aaaaaaaaaaaaaafq","aaaaaaaaaaaaaafr","aaaaaaaaaaaaaafs","aaaaaaaaaaaaaaft","aaaaaaaaaaaaaafu","aaaaaaaaaaaaaafv","aaaaaaaaaaaaaafw","aaaaaaaaaaaaaafx","aaaaaaaaaaaaaafy","aaaaaaaaaaaaaafz","aaaaaaaaaaaaaaga","aaaaaaaaaaaaaagb","aaaaaaaaaaaaaagc","aaaaaaaaaaaaaagd","aaaaaaaaaaaaaage","aaaaaaaaaaaaaagf","aaaaaaaaaaaaaagg","aaaaaaaaaaaaaagh","aaaaaaaaaaaaaagi","aaaaaaaaaaaaaagj","aaaaaaaaaaaaaagk","aaaaaaaaaaaaaagl","aaaaaaaaaaaaaagm","aaaaaaaaaaaaaagn","aaaaaaaaaaaaaago","aaaaaaaaaaaaaagp","aaaaaaaaaaaaaagq","aaaaaaaaaaaaaagr","aaaaaaaaaaaaaags","aaaaaaaaaaaaaagt","aaaaaaaaaaaaaagu","aaaaaaaaaaaaaagv","aaaaaaaaaaaaaagw","aaaaaaaaaaaaaagx","aaaaaaaaaaaaaagy","aaaaaaaaaaaaaagz","aaaaaaaaaaaaaaha","aaaaaaaaaaaaaahb","aaaaaaaaaaaaaahc","aaaaaaaaaaaaaahd","aaaaaaaaaaaaaahe","aaaaaaaaaaaaaahf","aaaaaaaaaaaaaahg","aaaaaaaaaaaaaahh","aaaaaaaaaaaaaahi","aaaaaaaaaaaaaahj","aaaaaaaaaaaaaahk","aaaaaaaaaaaaaahl","aaaaaaaaaaaaaahm","aaaaaaaaaaaaaahn","aaaaaaaaaaaaaaho","aaaaaaaaaaaaaahp","aaaaaaaaaaaaaahq","aaaaaaaaaaaaaahr","aaaaaaaaaaaaaahs","aaaaaaaaaaaaaaht","aaaaaaaaaaaaaahu","aaaaaaaaaaaaaahv","aaaaaaaaaaaaaahw","aaaaaaaaaaaaaahx","aaaaaaaaaaaaaahy","aaaaaaaaaaaaaahz","aaaaaaaaaaaaaaia","aaaaaaaaaaaaaaib","aaaaaaaaaaaaaaic","aaaaaaaaaaaaaaid","aaaaaaaaaaaaaaie","aaaaaaaaaaaaaaif","aaaaaaaaaaaaaaig","aaaaaaaaaaaaaaih","aaaaaaaaaaaaaaii","aaaaaaaaaaaaaaij","aaaaaaaaaaaaaaik","aaaaaaaaaaaaaail","aaaaaaaaaaaaaaim","aaaaaaaaaaaaaain","aaaaaaaaaaaaaaio","aaaaaaaaaaaaaaip","aaaaaaaaaaaaaaiq","aaaaaaaaaaaaaair","aaaaaaaaaaaaaais","aaaaaaaaaaaaaait","aaaaaaaaaaaaaaiu","aaaaaaaaaaaaaaiv","aaaaaaaaaaaaaaiw","aaaaaaaaaaaaaaix","aaaaaaaaaaaaaaiy","aaaaaaaaaaaaaaiz","aaaaaaaaaaaaaaja","aaaaaaaaaaaaaajb","aaaaaaaaaaaaaajc",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// Code
/*
class Solution {
public:
    int row,col;
    int targetk;
    string str;
    int dx[4]={-1,0,1,0};
    int dy[4]={0,-1,0,1};
    char used='#';
    vector< vector > grid;
    vector< vector > origin;
    vector< string> ans;
    bool flag;
    vector findWords(vector< vector >& board, vector& words)
    {
        row=board.size();
        if (row==0) return ans;
        col=board[0].size();
        grid.resize(row+2);
        for(int i=0;i<=col+1;i++)
            grid[0].push_back(used);
        for(int i=0;i tmpw;
        map repeat;
        for(int t=0;t

84 Largest Rectangle in Histogram

// -------------------- Seeing Discuss ----------
/*
	copying codes
	url
		https://discuss.leetcode.com/topic/14406/my-concise-code-20ms-stack-based-o-n-one-trick-used
	Analysis:
	The idea is simple, use a stack to save the index of each vector entry in a ascending order; 
	once the current entry is smaller than the one with the index s.top(), then that means the 
	rectangle with the height height[s.top()] ends at the current position, so calculate its area 
	and update the maximum.
	The only trick I use to avoid checking whether the stack is empty (due to pop) and also avoiding 
	emptying the stack at the end (i.e. after going through the vector, s is not empty and we have to 
	consider those in the stack) is to put a dummy "0" at the beginning of vector "height" and the end 
	of "height": the first one makes sure the stack will never be empty (since all the height entries are >=0) 
	and the last one will flush all the remaining non-zero entries of the stack at the end of "for" iteration. 
	This trick helps us keep the code concise.
*/
class Solution {
public:
    int largestRectangleArea(vector& heights)
    {
        heights.insert(heights.begin(),0);
        heights.push_back(0);
        int len = heights.size();
        int i ,res = 0,idx;
        stack s;
        s.push(0);
        for(i=1;i& heights)
    {
        int ans = 0;
        int lh = heights.size();
        int lhm = lh -1;
        int left,right;
        for(int i=0;i0 && heights[left]>=heights[i])
                left--;
            if (heights[left]=heights[i])
                right++;
            if (heights[right]ans) ans=nums;
        }
        return ans;
    }
};
*/

130 Surrounded Regions

// ----------------Seeing Discuss-----------
/*
	copying codes
	url
		https://discuss.leetcode.com/topic/17224/a-really-simple-and-readable-c-solution-only-cost-12ms
	Analysis:
		First, check the four border of the matrix. If there is a element is
		'O', alter it and all its neighbor 'O' elements to '1'.
		Then ,alter all the 'O' to 'X'
		At last,alter all the '1' to 'O'
		哦，我还说有要怎么处理边界上的O，原来一开始就把所有跟边界连着的O都标记出来。
		然后，最后再把这些标记过的O都变成O。
	小结：
		先把特殊的地方找出来单独标注，最后处理完所有后，再把特殊的地方覆盖上去。
		也是不错的办法。
*/
class Solution {
public:
    void solve(vector >& board)
    {
        int i,j;
        int row = board.size();
        if(!row) return ;
        int col=board[0].size();
        for(i=0;i1)
                check(board,i,col-1,row,col);
        }
        for(j=1;j+11)
                check(board,row-1,j,row,col);
        }
        for(i=0;i >&vec,int i,int j,int row,int col)
    {
        if(vec[i][j]=='O')
        {
            vec[i][j]='1';
            if(i>1) check(vec,i-1,j,row,col);
            if(j>1) check(vec,i,j-1,row,col);
            if(i+1

218 The Skyline Problem

// ------------------- Seeing Discuss ---------------
/*
	copying codes
	url
		https://discuss.leetcode.com/topic/25794/17-line-o-n-log-n-time-o-n-space-c-accepted-easy-solution-w-explanations
	
	Analysis:
	Step 1:
		Use a multimap to sort all boundary points. For a start point of an interval, 
		let the height be positive; otherwise, let the height be negative. Time complexity: O(n log n)
	Step 2:
		Use a multiset (rather than a heap/priority_queue) to maintain the current set of heights 
		to be considered. If a new start point is met, insert the height into the set, otherwise, 
		delete the height. The current max height is the back() element of the multiset. 
		For each point, the time complexity is O(log n). The overall time complexity is O(n log n).
	Step 3:
		Delete the points with equal heights. Time: O(n)
	Time Complexity: O(n log n)
	Space Complexity: O(n)
*/

class Solution {
public:
    vector > getSkyline(vector>& buildings)
    {
        multimap coords;
        for(const vector & building:buildings)
        {
            coords.emplace(building[0],building[2]);
            coords.emplace(building[1],-building[2]);
        }
        multiset heights ={0};
        vector > corners;
        int x = -1;
        int y = 0;
        for(const pair &p : coords)
        {
            if((x>=0)&&(p.first!=x)&&(corners.empty() || (corners.rbegin()->second !=y)))
            {
                corners.emplace_back(x,y);
            }
            if(p.second>=0)
            {
                heights.insert(p.second);
            }
            else
            {
                heights.erase(heights.find(-p.second));
            }
            x=p.first;
            y=*heights.rbegin();
        }
        if(!corners.empty()) corners.emplace_back(x,0);
        return corners;
    }
};