79 lines
3.3 KiB
Java
79 lines
3.3 KiB
Java
# 12. 矩阵中的路径
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[牛客网](https://www.nowcoder.com/practice/69fe7a584f0a445da1b6652978de5c38?tpId=13&tqId=11218&tab=answerKey&from=cyc_github)
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## 题目描述
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判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一个格子开始,每一步可以在矩阵中向上下左右移动一个格子。如果一条路径经过了矩阵中的某一个格子,则该路径不能再进入该格子。
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例如下面的矩阵包含了一条 bfce 路径。
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/1db1c7ea-0443-478b-8df9-7e33b1336cc4.png" width="200px"> </div><br>
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## 解题思路
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使用回溯法(backtracking)进行求解,它是一种暴力搜索方法,通过搜索所有可能的结果来求解问题。回溯法在一次搜索结束时需要进行回溯(回退),将这一次搜索过程中设置的状态进行清除,从而开始一次新的搜索过程。例如下图示例中,从 f 开始,下一步有 4 种搜索可能,如果先搜索 b,需要将 b 标记为已经使用,防止重复使用。在这一次搜索结束之后,需要将 b 的已经使用状态清除,并搜索 c。
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png" width="200px"> </div><br>
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本题的输入是数组而不是矩阵(二维数组),因此需要先将数组转换成矩阵。
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```java
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public class Solution {
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private final static int[][] next = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
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private int rows;
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private int cols;
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public boolean hasPath (String val, int rows, int cols, String path) {
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if (rows == 0 || cols == 0) return false;
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this.rows = rows;
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this.cols = cols;
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char[] array = val.toCharArray();
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char[][] matrix = buildMatrix(array);
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char[] pathList = path.toCharArray();
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boolean[][] marked = new boolean[rows][cols];
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for (int i = 0; i < rows; i++)
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for (int j = 0; j < cols; j++)
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if (backtracking(matrix, pathList, marked, 0, i, j))
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return true;
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return false;
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}
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private boolean backtracking(char[][] matrix, char[] pathList,
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boolean[][] marked, int pathLen, int r, int c) {
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if (pathLen == pathList.length) return true;
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if (r < 0 || r >= rows || c < 0 || c >= cols
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|| matrix[r][c] != pathList[pathLen] || marked[r][c]) {
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return false;
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}
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marked[r][c] = true;
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for (int[] n : next)
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if (backtracking(matrix, pathList, marked, pathLen + 1, r + n[0], c + n[1]))
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return true;
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marked[r][c] = false;
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return false;
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}
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private char[][] buildMatrix(char[] array) {
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char[][] matrix = new char[rows][cols];
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for (int r = 0, idx = 0; r < rows; r++)
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for (int c = 0; c < cols; c++)
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matrix[r][c] = array[idx++];
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return matrix;
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}
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public static void main(String[] args) {
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Solution solution = new Solution();
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String val = "ABCESFCSADEE";
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int rows = 3;
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int cols = 4;
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String path = "ABCCED";
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boolean res = solution.hasPath(val, rows, cols, path);
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System.out.println(res);
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}
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}
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```
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