【LeetCode】 63. Unique Paths II 不同路径 II（Medium）（JAVA）
题目地址: https://leetcode-.com/problems/unique-paths-ii/
题目描述：
A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
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.
Note: m and n will be at most 100.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right
题目大意
一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。
机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。
现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？
解题方法
1、采用动态规划，从后往前遍历
2、遇到 1，直接跳过，不然就把前和上的值相加
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid.length == 0 || obstacleGrid[0].length == 0 || obstacleGrid[obstacleGrid.length - 1][obstacleGrid[0].length - 1] == 1) return 0;
int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length];
dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1] = 1;
for (int i = obstacleGrid.length - 1; i >= 0; i--) {
for (int j = obstacleGrid[0].length - 1; j >= 0; j--) {
if (obstacleGrid[i][j] == 1) continue;
if (i == obstacleGrid.length - 1 && j == obstacleGrid[0].length - 1) {
dp[i][j] = 1;
} else if (i == obstacleGrid.length - 1) {
dp[i][j] = dp[i][j + 1];
} else if (j == obstacleGrid[0].length - 1) {
dp[i][j] = dp[i + 1][j];
} else {
dp[i][j] = dp[i][j + 1] + dp[i + 1][j];
}
}
}
return dp[0][0];
}
}
执行用时 : 1 ms, 在所有 Java 提交中击败了 83.38% 的用户
内存消耗 : 37.7 MB, 在所有 Java 提交中击败了 67.00% 的用户
