62.不同路径

题目
文章讲解
视频讲解

思路：确定dp[i][j]含义：到达[i,j]有多少条路径；递推公式：dp[i][j]=dp[i][j-1]+dp[i-1][j]

class Solution {public int uniquePaths(int m, int n) {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][j-1]+dp[i-1][j];}}return dp[m-1][n-1];}
}

63. 不同路径 II

题目
文章讲解
视频讲解

思路：在上一题的思路上增加一个障碍，存在障碍则dp[i][j]表示为0（初始状态）

class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[][] dp = new int[m][n];if (obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1) {return 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++) {dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;}}return dp[m - 1][n - 1];}
}