dp-基础版动态规划（动态规划每日一题计划）10/50

 

最小路径和

class Solution {public static int minPathSum(int[][] grid) {int dp[][]=new int[grid.length][grid[0].length];dp[0][0]=grid[0][0];for(int i=1;i<grid[0].length;i++){dp[0][i]=grid[0][i]+dp[0][i-1];}for(int i=1;i<grid.length;i++){dp[i][0]=grid[i][0]+dp[i-1][0];}for(int i=1;i< grid.length;i++){for(int j=1;j<grid[i].length;j++){dp[i][j]=Math.min(dp[i-1][j],dp[i][j-1])+grid[i][j];}}return dp[grid.length-1][grid[0].length-1];}
}

初始化第一行和第一列，除了第一行第一列，其他的每个位置继承上/左的距离，选择最短的那个即可。 

不同路径Ⅱ

import java.util.Arrays;
class Solution {public static int uniquePathsWithObstacles(int[][] obstacleGrid) {int dp[][] = new int[obstacleGrid.length][obstacleGrid[0].length];for (int i = 0; i < obstacleGrid.length; i++) {for (int j = 0; j < obstacleGrid[0].length; j++) {if (i == 0 && j == 0 && obstacleGrid[i][j] != 1) {dp[i][j] = 1;continue;}if (obstacleGrid[i][j] == 1) {dp[i][j] = 0;} else {try {dp[i][j] += dp[i - 1][j] + dp[i][j - 1];} catch (Exception e) {try {dp[i][j] += dp[i][j - 1];} catch (Exception k) {dp[i][j] += dp[i - 1][j];}}}}}return dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1];}
}

判断，嵌套...

三角形最小路径和

数塔问题

class Solution {public static int minimumTotal(List<List<Integer>> triangle) {int f[][] = new int[triangle.size()][triangle.get(triangle.size() - 1).size()];// 初始值 f[i][j] 4 1 8 3for (int i = 0; i < triangle.get(triangle.size() - 1).size(); i++) {f[triangle.size() - 1][i] = triangle.get(triangle.size() - 1).get(i);}for (int i = triangle.size() - 1; i >= 0; i--) {for (int j = 0; j < triangle.get(i).size(); j++) {try {f[i][j] += Math.min(f[i + 1][j], f[i + 1][j + 1]) + triangle.get(i).get(j);} catch (Exception e) {f[i][j] = triangle.get(i).get(j);}}}return f[0][0];}
}

下降路径最小和​​​​​​​ 

class Solution {public static void main(String[] args) {System.out.println(minFallingPathSum(new int[][]{{2, 1, 3}, {6, 5, 4}, {7, 8, 9}}));}public static int minFallingPathSum(int[][] matrix) {int f[][] = new int[matrix.length][matrix[0].length + 2];for (int i = 0; i < matrix.length; i++) {for (int j = 0; j < matrix[0].length + 2; j++) {f[i][j] = Integer.MAX_VALUE;}}for (int i = 1; i <= matrix[0].length; i++) {f[matrix.length - 1][i] = matrix[matrix.length - 1][i - 1];}for (int i = matrix.length - 2; i >= 0; i--) {for (int j = 1; j <= matrix[0].length; j++) {f[i][j] = Math.min(f[i + 1][j], Math.min(f[i + 1][j + 1], f[i + 1][j - 1])) + matrix[i][j - 1];}}int minv = Integer.MAX_VALUE;for (int i = 1; i <= matrix[0].length; i++) {minv = Math.min(minv, f[0][i]);}return minv;}
}