Valid Sudoku & Subsets I&&II & Flatten Binary Tree to Linked List & Jump Game I&&II

(1) Valid Sudoku

就是每一行，每一列，每一个方块的check，还有一种简化版[1]:

class Solution {public:    bool isValidSudoku(vector<vector<char> > &board) {        int row[9],col[9],area[9];                for(int i=0;i<9;i++)        {            memset(row,0,9*sizeof(int));            memset(col,0,9*sizeof(int));                        for(int j=0;j<9;j++)            {                if(board[i][j]!='.')                {                    if(row[board[i][j]-'1']>0)                        return false;                    else                        row[board[i][j]-'1']++;                }                                if(board[j][i]!='.')                {                    if(col[board[j][i]-'1']>0)                        return false;                    else                        col[board[j][i]-'1']++;                }            }        }                for(int i=0;i<9;i+=3)            for(int j=0;j<9;j+=3)            {                memset(area,0,9*sizeof(int));                for(int k=0;k<3;k++)                    for(int l=0;l<3;l++)                        if(board[i+k][j+l]!='.')                        {                            if(area[board[i+k][j+l]-'1']>0)                                return false;                            else                                area[board[i+k][j+l]-'1']++;                        }            }        return true;    }};

(2) Subsets

DFS每次返回当前解，当解深度大于最大深度时返回：

class Solution {private:    void solve(int dep, int maxDep, vector<int> tmp,vector<int> S,vector<vector<int>> &ret, int begin){        ret.push_back(tmp);        if(dep==maxDep)            return;        for(int i=begin;i<S.size();i++)        {            vector<int> a(tmp);            a.push_back(S[i]);            solve(dep+1,maxDep,a,S,ret,i+1);        }    }public:    vector<vector<int> > subsets(vector<int> &S) {        vector<vector<int>> ret;        vector<int> tmp;        tmp.clear();        sort(S.begin(),S.end());        solve(0,S.size(),tmp,S,ret,0);        return ret;    }};

(3) Subset II

比(2)多一个检查是否重复的判断：

class Solution {private:    void solve(int dep, int maxDep, vector<int> tmp,vector<int> S,vector<vector<int>> &ret, int begin){        ret.push_back(tmp);        if(dep==maxDep)            return;        for(int i=begin;i<S.size();i++)        {            vector<int> a(tmp);                        if(i>begin)            {                if(S[i]>S[i-1])                {                    a.push_back(S[i]);                    solve(dep+1,maxDep,a,S,ret,i+1);                }            }            else            {                a.push_back(S[i]);                solve(dep+1,maxDep,a,S,ret,i+1);            }        }    }public:    vector<vector<int> > subsetsWithDup(vector<int> &S) {        vector<vector<int>> ret;        vector<int> tmp;        tmp.clear();        sort(S.begin(),S.end());        solve(0,S.size(),tmp,S,ret,0);        return ret;    }};

(4) Flatten Binary Tree to Linked List

非递归后序遍历：

class Solution {public:    void flatten(TreeNode *root) {        if(!root)            return;        stack<TreeNode *> stk;        TreeNode *preNode=NULL,*curNode=NULL;          stk.push(root);                while(!stk.empty())        {            curNode=stk.top();            stk.pop();                        if(preNode)            {                preNode->left=NULL;                preNode->right=curNode;            }                        preNode=curNode;            if(curNode->right)                stk.push(curNode->right);            if(curNode->left)                stk.push(curNode->left);        }    }};

(5) Jump Game 

只用关心最远能跳到什么地方[2]:

class Solution {public:    bool canJump(int A[], int n) {        if(n==0)            return false;        int maxPos=A[0];                for(int i=0;i<=min(maxPos,n-1);i++)            maxPos=max(maxPos,i+A[i]);                return maxPos>=n-1;    }};

(6) Jump Game II

根据[3]：

class Solution {public:    int jump(int A[], int n) {        int jump[n];        int maxJump=0;                memset(jump,0,n*sizeof(int));                for(int i=0;i<n;i++)        {            for(int j=maxJump-i+1;j<=A[i] && i+j<n;j++)                jump[i+j]=jump[i]+1;                        maxJump=max(maxJump,i+A[i]);        }        return jump[n-1];    }};

参考：

[1]http://blog.csdn.net/doc_sgl/article/details/13002461

[2]http://blog.csdn.net/pickless/article/details/9855891

[3]http://blog.csdn.net/pickless/article/details/9776661