63. Unique Paths II

大意同62，m*n的矩阵，但有一些位置上存储着1，表示不能走

思路：动态规划

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class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        # 6 star, 动态规划， 注意要特殊处理最上和左的行和列
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        dp = [[0 for _ in range(n)] for _ in range(m)]
        for i in range(m):
            if obstacleGrid[i][0] == 0:
                dp[i][0] = 1
            else:
                break
        for j in range(n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = 1
            else:
                break

        for i in range(1, n):
            for j in range(1, m):
                if obstacleGrid[j][i]:
                    dp[j][i] = 0
                else:
                    dp[j][i] = dp[j-1][i] + dp[j][i-1]
        return dp[-1][-1]

Go:

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func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	if obstacleGrid[0][0] == 1 {return 0}
	cols := len(obstacleGrid)
	rows := len(obstacleGrid[0])
	dp := make([][]int, cols)
	for i:=0; i<cols; i++ {
		dp[i] = make([]int, rows)
		if i == 0 {dp[i][0] = 1}
		if (i > 0 && dp[i-1][0] == 0) || obstacleGrid[i][0] == 1 {
			dp[i][0] = 0
		} else {
			dp[i][0] = 1
		}

	}
	for j:=0; j<rows; j++ {
		if (j > 0 && dp[0][j-1] == 0) || obstacleGrid[0][j] == 1 {
			dp[0][j] = 0
		} else {
			dp[0][j] = 1
		}
	}
	for i:=1;i<cols;i++ {
		for j:=1; j<rows; j++ {
			if obstacleGrid[i][j] == 1{
				dp[i][j] = 0
			} else {
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}

		}
	}
	return dp[cols-1][rows-1]

}