1. 确定dp数组（dp table）以及下标的含义
2. 确定递推公式
3. dp数组如何初始化
4. 确定遍历顺序
5. 举例推导dp数组

62.不同路径 

代码随想录

视频讲解：动态规划中如何初始化很重要！| LeetCode：62.不同路径_哔哩哔哩_bilibili

class Solution {public int uniquePaths(int m, int n) {// 1.确定dp数组（dp table）以及下标的含义 : dp[i][j] 从[0][0]到[i][j]有多少种路径(是路径不是步数)// 2.确定递推公式 : dp[i][j]只能由这两个方向的路径推导出来dp[i - 1][j],dp[i][j - 1]// dp[i][j] = dp[i - 1][j] + dp[i][j - 1]// 3.dp数组如何初始化 : dp[0][j] = 1; dp[i][0] = 1; 画图，只有一种路径// 4.确定遍历顺序 : 从左往右，从上到下// 5.举例推导dp数组 :int dp[][] = new int[m][n];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];}
}

 63. 不同路径 II 

https://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.htmlhttps://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.html

视频讲解：动态规划，这次遇到障碍了| LeetCode：63. 不同路径 II_哔哩哔哩_bilibili

class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int dp[][] = new int[obstacleGrid.length][obstacleGrid[0].length];for (int i = 0; i < obstacleGrid.length && obstacleGrid[i][0] == 0; i++) {dp[i][0] = 1;}for (int j = 0; j < obstacleGrid[0].length && obstacleGrid[0][j] == 0; j++) {dp[0][j] = 1;}for (int i = 1; i < obstacleGrid.length; i++) {for (int j = 1; j < obstacleGrid[0].length; j++) {if (obstacleGrid[i][j] == 0) {dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}}return dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1];}
}