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T65.java
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49 lines (46 loc) · 2.28 KB
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package web; /**
* @program LeetNiu
* @description: 矩阵中的路径
* @author: mf
* @create: 2020/01/17 23:43
*/
/**
* 请设计一个函数,用来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左,向右,向上,向下移动一个格子。
* 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。
* 例如 a b c e s f c s a d e e 矩阵中包含一条字符串"bcced"的路径,
* 但是矩阵中不包含"abcb"路径,因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后,路径不能再次进入该格子。
*/
public class T65 {
public boolean hasPath(char[] matrix, int rows, int cols, char[] str) {
if (null == matrix || matrix.length == 0 || null == str || str.length == 0 || matrix.length != rows * cols || rows <= 0 || cols <= 0){
return false;
}
boolean[] visited = new boolean[rows * cols];
int[] pathLength = {0};
for (int i = 0; i <= rows - 1; i++){
for (int j = 0; j <= cols - 1; j++){
if (hasPathCore(matrix, rows, cols, str, i, j, visited, pathLength)) {
return true;
}
}
}
return false;
}
public boolean hasPathCore(char[] matrix, int rows, int cols, char[] str,int row,int col, boolean[] visited,int[] pathLength) {
boolean flag = false;
if (row >= 0 && row < rows && col >= 0 && col < cols && !visited[row * cols + col] && matrix[row * cols + col] == str[pathLength[0]]) {
pathLength[0]++;
visited[row * cols + col] = true;
if (pathLength[0] == str.length) {
return true;
}
flag = hasPathCore(matrix, rows, cols, str, row, col + 1, visited, pathLength) || hasPathCore(matrix, rows, cols, str, row + 1, col, visited, pathLength) || hasPathCore(matrix, rows, cols, str, row, col - 1, visited, pathLength) || hasPathCore(matrix, rows, cols, str, row - 1, col, visited, pathLength);
if (!flag) {
pathLength[0]--;
visited[row * cols + col] = false;
}
}
return flag;
}
}