算法：回溯法 矩阵中的路径

题目：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格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子
利用回溯法：直接上代码

	public static boolean hasPath(char[][] matrix,char[] str){
		int rows= matrix.length;
		int cols = matrix[0].length;
		boolean[][] visited = new boolean[rows][cols];
		int pathLength = 0;
		
		for(int row = 0 ; row < rows ; row++){
			for(int col = 0 ; col < cols ; col++){
				if(hasPathCore(matrix,rows,cols,row,col,str,pathLength,visited)){
					return true;
				}
			}
		}
		return false;
	}

	private static boolean hasPathCore(char[][] matrix, int rows, int cols, int row, int col, char[] str,
			int pathLength, boolean[][] visited) {
			boolean hasPath = false;
			if(row > 0 &&row < rows &&col > 0 && col < cols && matrix[row][col] == str[pathLength] && !visited[row][col]){
				pathLength ++;
				visited[row][col] = true;
				hasPath = hasPathCore(matrix,rows, cols, row-1, col, str,pathLength, visited)
						||hasPathCore(matrix,rows, cols, row, col-1, str,pathLength, visited)
						||hasPathCore(matrix,rows, cols, row+1, col, str,pathLength, visited)
						||hasPathCore(matrix,rows, cols, row, col+1, str,pathLength, visited);
				if(!hasPath){
					pathLength--;
					visited[row][col] = false;
				}
			}
		return hasPath;
	}

								***帅气的远远啊***