不同路径I

思路：定义一个二维的dp数组

1.dp[i][j]表示从（0，0）到（i，j）的路径条数

2.dp[i][j] = dp[i-1][j] + dp[i][j-1];从上一状态得到当前状态方程

两个方向过来当前位置，是两条路径，路径条数，不是步数

3.初始化

dp[i][0] = 1;

dp[0][j] = 1;

4.确定遍历顺序

从上到下 或 从左到右

5.举例推导，尝试带入 3，7，手动打印

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

优化：

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

 

不同路径II

思路分析：加入了障碍，标记对应的dp table，保持初始值（0）即可。

1.dp[i][j]表示从（0，0）到（i，j）的不同路径数

2.dp[i][j] = dp[i - 1][j] + dp[i][j -1].

如果（i,j）就是障碍的话，应该保持初始状态（初始状态为0）

if (obstacleGrid[i][j] == 0) { // 当(i, j)没有障碍的时候，再推导dp[i][j]dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}

3.初始化

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;

obs为0时才进行，为1不进行

4.遍历顺序：从左到右，从上到下

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

5.打印debug

class Solution{
public:int uniquePathWithObtacles(vector<vector<int>>&obsracleGrid){int m = obstacleGrid.size();int n = obstacleGrid[0].size();if(obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1)//终点或起点出现障碍return 0;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] == 1)continue;dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}return dp[m - 1][n - 1];}
};

优化：

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