0063. Unique Paths I I

63. Unique Paths II #

题目 #

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.

Notem 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?

解题思路 #

  • 这一题是第 62 题的加强版。也是一道考察 DP 的简单题。
  • 这一题比第 62 题增加的条件是地图中会出现障碍物,障碍物的处理方法是 dp[i][j]=0
  • 需要注意的一种情况是,起点就是障碍物,那么这种情况直接输出 0 。

代码 #


package leetcode

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	if len(obstacleGrid) == 0 || obstacleGrid[0][0] == 1 {
		return 0
	}
	m, n := len(obstacleGrid), len(obstacleGrid[0])
	dp := make([][]int, m)
	for i := 0; i < m; i++ {
		dp[i] = make([]int, n)
	}
	dp[0][0] = 1
	for i := 1; i < n; i++ {
		if dp[0][i-1] != 0 && obstacleGrid[0][i] != 1 {
			dp[0][i] = 1
		}
	}
	for i := 1; i < m; i++ {
		if dp[i-1][0] != 0 && obstacleGrid[i][0] != 1 {
			dp[i][0] = 1
		}
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if obstacleGrid[i][j] != 1 {
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}
		}
	}
	return dp[m-1][n-1]
}


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