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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/
接着上一题。
http://www.cnblogs.com/Liok3187/p/4646924.html
还是递推，当前格值就是上方的值加上左边的值。
每一轮都做一个"厂"字，以左上为中心，先往右再往下。
处理障碍:在往右和往下的途中，如果遇到了障碍，这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
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原文地址：http://www.cnblogs.com/Liok3187/p/4668974.html
