代码随想录Day39

Day 39 动态规划 part02

今日任务

62.不同路径
1. 不同路径 II

代码实现

62.不同路径

    public int uniquePaths(int m, int n) {if (m <= 1 || n<= 1) return 1;int[][] dp = new int[m][n];for (int i = 0; i < dp.length; i++) {for (int j = 0; j < dp[i].length; j++) {if (i == 0 || j == 0) {dp[i][j] = 1;} else {dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}}return dp[m - 1][n - 1];}

不同路径 II

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {if (obstacleGrid[0][0] == 1) return 0;int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length];for (int i = 0; i < dp.length; i++) {for (int j = 0; j < dp[i].length; j++) {if (obstacleGrid[i][j] == 1) {dp[i][j] = 0;} else if (i == 0) {if (j == 0) {dp[i][j] = 1;} else {dp[i][j] = Math.min(dp[i][j - 1], 1);}} else if (j == 0) {dp[i][j] = Math.min(dp[i - 1][j], 1);} else {dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}}return dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1];}
//代码随想录版本，代码简洁一些
class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[][] dp = new int[m][n];//如果在起点或终点出现了障碍，直接返回0if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 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];}
}