unique-paths-ii

最新推荐文章于 2019-06-19 21:23:05 发布

原创最新推荐文章于 2019-06-19 21:23:05 发布 · 154 阅读

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本文介绍了一种算法，用于计算在存在障碍物的情况下从起点到终点的不同路径数量。该算法使用动态规划方法，通过构建二维数组来存储到达每个点的路径数量，并考虑障碍物的影响。

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题目描述

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 as1and0respectively 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 is2.

Note: m and n will be at most 100.

这道题也不难

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        	int m = obstacleGrid.length;
		int n = obstacleGrid[0].length;
		int[][] dep = new int[m][n];
		if(obstacleGrid[0][0]==1)return 0;
		dep[0][0] = 1;
		boolean o = false;
		for (int i = 0; i < m; i++) {
			if(o) {
				dep[i][0]=0;
			}else {
				if(obstacleGrid[i][0]==1) {
					o=true;
					dep[i][0]=0;
				}else {
					dep[i][0]=1;
				}
			}
		}
		o=false;
		for (int i = 0; i < n; i++) {
			if(o) {
				dep[0][i]=0;
			}else {
				if(obstacleGrid[0][i]==1) {
					o=true;
					dep[0][i]=0;
				}else {
					dep[0][i]=1;
				}
			}
		}
		for (int i = 1; i < m; i++) {
			for (int j = 1; j < n; j++) {
				if(obstacleGrid[i][j]==1) {
					dep[i][j] =0;
				}else {
					dep[i][j]=dep[i-1][j]+dep[i][j-1];
				}
			}
		}
		return dep[m-1][n-1];
    }
}