《程序代码面试指南》 已完成 递归和动态规划-矩阵的最小路径和

Author: 233zzh

java/zuochengyun/recursion_dp/minPathSumOfMatrix.java

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+package zuochengyun.recursion_dp;
+
+/**
+ * @Description 矩阵的最小路径和 https://leetcode.cn/problems/0i0mDW/
+ * @Author 爱做梦的鱼
+ * @Blog https://zihao.blog.csdn.net/
+ * @Date 2023/5/17 16:21
+ */
+public class minPathSumOfMatrix {
+
+  public int getMinPathSum(int[][] matrix) {
+    int m = matrix.length;
+    int n = matrix[0].length;
+    int[][] dp = new int[m][n];
+    dp[0][0] = matrix[0][0];
+    for (int i = 1; i < m; i++) {
+      dp[i][0] = dp[i - 1][0] + matrix[i][0];
+    }
+    for (int j = 1; j < n; j++) {
+      dp[0][j] = dp[0][j - 1] + matrix[0][j];
+    }
+    for (int i = 1; i < m; i++) {
+      for (int j = 1; j < n; j++) {
+        dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + matrix[i][j];
+      }
+    }
+    return dp[m - 1][n - 1];
+  }
+
+  //给定一个矩阵 m，从左上角开始每次只能向右或者向下走，最后到达右下角的位置，路径上所有的数字累加起来就是路径和，返回所有的路径中最小的路径和。
+  // 贪心算法
+  public int getMinPathSum1(int[][] matrix) {
+    return dfs(matrix, 0, 0, matrix[0][0]);
+  }
+
+  public int dfs(int[][] matrix, int i, int j, int result) {
+    System.out.println(i + " " + j);
+    int m = matrix.length;
+    int n = matrix[0].length;
+    if (i == m - 1 && j == n - 1) {
+      return result;
+    }
+    if (i == m - 1) {
+      return dfs(matrix, i, j + 1, result + matrix[i][j + 1]);
+    }
+    if (j == n - 1) {
+      return dfs(matrix, i + 1, j, result + matrix[i + 1][j]);
+    }
+    if (matrix[i + 1][j] < matrix[i][j + 1]) {
+      return dfs(matrix, i + 1, j, result + matrix[i + 1][j]);
+    } else {
+      return dfs(matrix, i, j + 1, result + matrix[i][j + 1]);
+    }
+  }
+
+  public static void main(String[] args) {
+    int[][] matrix = new int[][]{{1, 3, 5, 9}, {8, 1, 3, 4}, {5, 0, 6, 1}, {8, 8, 4, 0}};
+    int result = new minPathSumOfMatrix().getMinPathSum(matrix);
+    System.out.println(result);
+  }
+
+}