【第39天】路径规划，看看即可

1.Unique Paths

Every once in a while 我需要找找思路看这题怎么做。

class Solution {
    public int uniquePaths(int m, int n) {
        int[][]dp = new int[m][n];

        for(int i = 0; i < dp.length; ++i){
            for(int j = 0; j < dp[0].length; ++j){
                if(i == 0 || j == 0){
                    dp[i][j] = 1;
                }else{
                    dp[i][j] = dp[i-1][j] + dp[i][j-1];
                }
            }
        }

        return dp[m-1][n-1];
    }
}

2.Unique Paths II

这题重在preprocessing。把第一行和第一列的元素process：如果有obstacle，就停止赋值1。那行/列后面的元素就自然是0了。
后面process计算的时候，跳过这个块即可。

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1) return 0;
        
        int[][] dp = new int[m][n];

        for(int i = 0; i < n && obstacleGrid[0][i] == 0; ++i) dp[0][i] = 1;
        for(int j = 0; j < m && obstacleGrid[j][0] == 0; ++j) dp[j][0] = 1;

        for(int i = 1; i < m; ++i){
            for(int j = 1; j < n; ++j){
                if(obstacleGrid[i][j] == 1){
                    continue;
                }else{
                    if(i == 0 || j == 0){
                        dp[i][j] = 1;
                    }else{
                        dp[i][j] = dp[i-1][j] + dp[i][j-1];
                    }
                }
            }
        }
        return dp[m-1][n-1];
    }
}