[LeetCode-interview algorithm classic-Java implementation] [063-Unique Paths II (Unique path Question II)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [063-Unique Paths II (Unique path Question II)], leetcode -- java
[063-Unique Paths II (Unique path Problem II )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

Follow up for "Unique Paths ":
Now consider if some obstacles are added to the grids. How many unique paths wocould 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 partitioned strated below.

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

The total number of unique paths is2.
Note: m and n will be at most100.

Theme

If there is an obstacle in the path, find the total number of paths. [Unique path problem]
　　Note:The number of rows and columns in the grid cannot exceed 100

Solutions

Divide and conquer
Save the result with m * n group.
For elements in array.
1. When x = 0 or y = 0, and (x, y) is not accessible. A [x] [y] = 1; 0
2. When x> = 1 and y> = 1, and (x, y) is not accessible. There is A [x] [y] = A [x-1] [y] + A [x] [Y-1]. The obstacle is 0.
3. The requested node is A [s-1] [n-1].

Code Implementation

Algorithm Implementation class

Public class Solution {public int uniquePathsWithObstacles (int [] [] obstacleGrid) {// input validation if (obstacleGrid = null | obstacleGrid. length <1 | obstacleGrid [0]. length <1 | obstacleGrid [0] [0] = 1 | obstacleGrid [obstacleGrid. length-1] [obstacleGrid [0]. length-1] = 1) {return 0;} int rows = obstacleGrid. length; int cols = obstacleGrid [0]. length; int [] [] result = new int [rows] [cols]; // The first How many methods are there for location? One is accessibility, and the other is 0 results [0] [0] = obstacleGrid [0] [0] = 0? 1: 0; for (int I = 1; I <cols; I ++) {result [0] [I] = obstacleGrid [0] [I] = 0? Result [0] [I-1]: 0;} for (int I = 1; I <rows; I ++) {result [I] [0] = obstacleGrid [I] [0] = 0? Result [I-1] [0]: 0 ;}for (int I = 1; I <rows; I ++) {for (int j = 1; j <cols; j ++) {result [I] [j] = obstacleGrid [I] [j] = 0? Result [I-1] [j] + result [I] [j-1]: 0;} return result [rows-1] [cols-1];} // recursive method will time out public int uniquePathsWithObstacles2 (int [] [] obstacleGrid) {// input validation if (obstacleGrid = null | obstacleGrid. length <1 | obstacleGrid [0]. length <1 | obstacleGrid [obstacleGrid. length-1] [obstacleGrid [0]. length-1] = 1) {return 0;} int [] result = {0}; solve (obstacleGrid, 0, 0, result); return result [0];} public void solve (int [] [] grid, int row, int col, int [] sum) {// if (row = grid. length-1 & col = grid [0]. length-1) {sum [0] ++;} // No end point exists. The current position is inside the chessboard and is not else if (row> = 0 & row <grid. length & col> = 0 & col <grid [0]. length & grid [row] [col] = 0) {// go to the Right to solve (grid, row, col + 1, sum); // go down to solve (grid, row + 1, col, sum );}}}

Evaluation Result

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Note Please refer to the following link for more information: http://blog.csdn.net/derrantcm/article/details/4252723]

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