leetcode Unique Paths II

来源：互联网发布：微三云源码编辑：程序博客网时间：2024/05/29 09:20

递归算法超时

超时的代码

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {        int iRow = obstacleGrid.size();        int iCol = obstacleGrid[0].size();                if(iRow==0||iCol==0)            return 0;                pathNum = 0;                uniquePathsWithObstaclesHelper(obstacleGrid, 0, 0, iRow-1, iCol-1);            }          void uniquePathsWithObstaclesHelper(vector<vector<int> > obstacleGrid, int iCur, int jCur, int finishRow, int finishCol)     {                 if((iCur==finishRow&&jCur==finishCol-1)||(jCur==finishCol&&iCur==finishRow-1))          {              pathNum++;              return;          }                   if(iCur<finishRow&&obstacleGrid[iCur+1][jCur]!=1)               uniquePathsWithObstaclesHelper(obstacleGrid, iCur+1, jCur, finishRow, finishCol);          if(jCur<finishCol&&obstacleGrid[iCur][jCur+1]!=1)               uniquePathsWithObstaclesHelper(obstacleGrid, iCur, jCur+1, finishRow, finishCol);                }         private:    int pathNum;};

动态规划

Accepted代码

与Unique Paths 想法一样，不过需要对障碍进行额外处理

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {        int m= obstacleGrid.size();        int n = obstacleGrid[0].size();                if(m==0||n==0)            return 0;        vector<vector<int> > pathNum(m, vector<int>(n,0));                if(obstacleGrid[0][0])            pathNum[0][0] = 0;        else            pathNum[0][0] = 1;                for(int i = 1; i < m; ++i)            pathNum[i][0] =(pathNum[i-1][0]&&!obstacleGrid[i][0]);        for(int j = 1; j < n; ++j)            pathNum[0][j] =(pathNum[0][j-1]&&!obstacleGrid[0][j]);                for(int i = 1; i < m; ++i)            for(int j = 1; j < n; ++j)            {                if(!obstacleGrid[i][j])                    pathNum[i][j] = pathNum[i-1][j]+pathNum[i][j-1];                else                    pathNum[i][j] = 0;            }                        return pathNum[m-1][n-1];    }};

0 0