[LeetCode][JavaScript]Unique Paths II

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Unique Paths II 

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[  [0,0,0],  [0,1,0],  [0,0,0]]

The total number of unique paths is 2.

Note: m and n will be at most 100.

https://leetcode.com/problems/unique-paths-ii/

 

 

 

 

接著上一題。

http://www.cnblogs.com/Liok3187/p/4646924.html

還是遞推，當前格值就是上方的值加上左邊的值。

每一輪都做一個"廠"字，以左上為中心，先往右再往下。

處理障礙:在往右和往下的途中，如果遇到了障礙，這一行或者一列就都是0了。

 1 /** 2  * @param {number[][]} obstacleGrid 3  * @return {number} 4  */ 5 var uniquePathsWithObstacles = function(obstacleGrid) { 6     var i, j, map = []; 7     var m = obstacleGrid.length, n = obstacleGrid[0].length, len = Math.min(m, n); 8     for(i = 0; i < m; i++){ 9         map[i] = [];10     }11 12     for(i = 0; i < len; i++){13         var meetOb = false;14         for(j = i; j < n; j++){15             if(obstacleGrid[i][j] === 1){16                 meetOb = true;17                 map[i][j] = 0;18                 continue;19             }20             if(i === 0 && !meetOb){21                 map[i][j] = 1;22             }else if(i === 0 && meetOb){23                 map[i][j] = 0;24             }else{25                 map[i][j] = map[i - 1][j] + map[i][j - 1];26             }27         }28         meetOb = false;29         for(j = i + 1; j < m; j++){30             if(obstacleGrid[j][i] === 1){31                 map[j][i] = 0;32                 meetOb = true;33                 continue;34             }35             if(i === 0 && !meetOb){36                 map[j][i] = 1;37             }else if(i === 0 && meetOb){38                 map[j][i] = 0;39             }else{40                 map[j][i] = map[j - 1][i] + map[j][i - 1];41             }42         }43     }44     return obstacleGrid[0][0] !== 1 ? map[m - 1][n - 1] : 0;45 };

 

 

 

 

[LeetCode][JavaScript]Unique Paths II