UniquePaths_2_63

https://leetcode-cn.com/problems/unique-paths-ii/

image.png

（图片来源https://leetcode-cn.com/problems/unique-paths-ii/

）

日期	是否一次通过	comment
2020-02-28	0	忘了处理grid边缘有1的情况

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初始化；交换；排障

1. row 向

   public int uniquePathsWithObstacles1(int[][] obstacleGrid) {
        int rowL = obstacleGrid.length;
        int colL = obstacleGrid[0].length;

        int[] dp = new int[rowL];
        // 初始化
        for(int i=0; i<rowL; i++) {
            if(obstacleGrid[i][0] != 0) {
                dp[i] = 0;
                break;
            }

            dp[i] = 1;
        }

        // 遍历grid
        for (int j=1; j<colL; j++) {
            if(obstacleGrid[0][j] != 0) {  // grid边缘处理
                dp[0] = 0;
            }
            for (int i=1; i < rowL; i++) {
                if (obstacleGrid[i][j] != 0) {
                    dp[i] = 0;
                } else { // j > 0 if (dp[i - 1] != 0)
                    dp[i] += dp[i - 1];
                }
            }
        }

        return dp[rowL - 1];
    }

2. col向

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int rowL = obstacleGrid.length;
        int colL = obstacleGrid[0].length;

        int[] dp = new int[colL];
        for(int j=0; j<colL; j++) {
            if(obstacleGrid[0][j] != 0) {
                dp[j] = 0;
                break;
            }

            dp[j] = 1;
        }


        for (int i=1; i<rowL; i++) {
            if(obstacleGrid[i][0] != 0) {
                dp[0] = 0;
            }
            for (int j=1; j < colL; j++) {
                if (obstacleGrid[i][j] != 0) {
                    dp[j] = 0;
                } else { // j > 0
                    dp[j] += dp[j - 1];
                }
            }
        }

        return dp[colL - 1];
    }