Dynamic Programming moderate leetcode 62. different paths

class Solution {
    public int uniquePaths(int m, int n) {
        /*
        dp[i][j]表示到达[i,j]方格有多少路径到达。
        一个方格的路径等于其上面方格和左面放个各自路径的和,即dp[i][j]=dp[i-1][j]+dp[i][j-1]
        第一行,第一列是特殊情况,
        basecase是dp[0][0]=1;
        */
        int [][]dp=new int[n][m];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(i==0&&j==0){
                    dp[i][j]=1;
                    continue;
                }
                if(i==0&&j!=0){
                    dp[i][j]=dp[i][j-1];
                    continue;
                }
                if(j==0&&i!=0){
                    dp[i][j]=dp[i-1][j];
                    continue;
                }
                dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[n-1][m-1];
    }
}
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Origin blog.csdn.net/weixin_43260719/article/details/104709437