LeetCode-63. Unique Paths II-优快云博客

本文介绍了一个算法问题，即在一个存在障碍物的网格中寻找从起点到终点的不同路径数量。通过修改原始的“独特路径”问题，加入障碍物，并提供了解决方案的代码实现。

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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.

这道题和上一题十分相像。只不过加了几个障碍。只需要把障碍所在的点置为0即可。所需注意的就是障碍在最上和最左的边上以及起始点的情况。代码如下：

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