62.不同路径

• dp[i][j]是从（0, 0）出发走到(i, j)的不同路径数.

• 限定了只能向右或者向下走，dp[i][j] = dp[i - 1][j] + dp[i][j - 1]，初始化边界的路径数都为1，遍历顺序从左上到右下每层遍历。

• 注意小trick，结果是dp[m - 1][n - 1]。自己想的，棒，dp找递推可以先逆着想。

//C++
class Solution {
public:
    int uniquePaths(int m, int n) {
        //vector<vector<int>> dp(m, vector<int>(n, 0));
        int dp[101][101];
        for (int i = 0; i < n; i++) {
            dp[0][i] = 1;
        }
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
            }
        }
        return dp[m - 1][n - 1];

    }
};

63. 不同路径 II

与上一题类似，只不过遇到障碍物保持dp为初始0，也就是跳过计算，同时初始化值的时候边界遇到障碍物后面dp也都维持0。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 0));
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
            dp[0][j] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 0)
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};